NCERT Exemplar for Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations (Book Solutions)

Solved Examples

Short Answer Type

Example 1. Evaluate ${(1 + i)^6} + {(1 - i)^3}$

Ans: Given ${(1 + i)^6} + {(1 - i)^3}$

The value of $i = \sqrt { - 1} {\text{, }}{i^2} =  - 1,{i^3} =  - i{\text{ and }}{i^4} = 1$

  ${(1 + i)^6} = {\left\{ {{{(1 + i)}^2}} \right\}^3}$

 $\Rightarrow {\left( {1 + {i^2} + 2i} \right)^3}$

 $\Rightarrow {(1 - 1 + 2i)^3}$

 $\Rightarrow 8{i^3} =  - 8i\left[ {\because {i^3} =  - i} \right]$ 

and 

  ${(1 - i)^3} = 1 - {i^3} - 3i + 3{i^2}$

 $\Rightarrow 1 + i - 3i - 3 =  - 2 - 2i$ 

Therefore, 

  ${(1 + i)^6} + {(1 - i)^3}$

 $\Rightarrow  - 8i - 2 - 2i =  - 2 - 10i$ 

Example 2. If ${(x + iy)^{\frac{1}{3}}} = a + ib$, where $x,y,a,b\in R$, show that $\dfrac{x}{a} - \dfrac{y}{b} =  - 2\left( {{a^2} + {b^2}} \right)$

Ans: Given, ${(x + iy)^{\frac{1}{3}}} = a + ib$, where $x,y,a,b \in {\text{R}}$

The value of $i = \sqrt { - 1} {\text{, }}{i^2} =  - 1,{i^3} =  - i{\text{ and }}{i^4} = 1$

${(x + iy)^{\frac{1}{3}}} = a + ib$

$\Rightarrow x + iy = {(a + ib)^3}$

i.e., $ x + iy = {a^3} + {i^3}{b^3} + 3iab(a + ib)$

$= {a^3} - i{b^3} + i3{a^2}b - 3a{b^2}$

$  = {a^3} - 3a{b^2} + i\left( {3{a^2}b - {b^3}} \right)$

On comparing real and imaginary parts, 

$\Rightarrow  x = {a^3} - 3a{b^2}$ and $y = 3{a^2}b - {b^3}$

Thus $ \dfrac{x}{a} = {a^2} - 3{b^2}$ and $\dfrac{y}{b} = 3{a^2} - {b^2}$

So, 

$\dfrac{x}{a} - \dfrac{y}{b} = {a^2} - 3{b^2} - 3{a^2} + {b^2}$

 $\Rightarrow  - 2{a^2} - 2{b^2} =  - 2\left( {{a^2} + {b^2}} \right).$ 

Example 3. Solve the equation ${z^2} = \bar z$, where $z = x + iy$.

Ans: Given, the equation ${z^2} = \bar z$, where $z = x + iy$.

The value of $i = \sqrt { - 1} {\text{, }}{i^2} =  - 1,{i^3} =  - i{\text{ and }}{i^4} = 1$

  ${z^2} = \bar z\quad$

 $\Rightarrow {x^2} - {y^2} + i2xy = x - iy$ 

Therefore, 

${x^2} - {y^2} = x  \ldots \left( 1 \right)$

and 

$2xy =  - y \ldots \left( 2 \right)$

From (2), we have $y = 0$ or $x =  - \dfrac{1}{2}$

When $y = 0$, from (1), we get ${x^2} - x = 0$, i.e., $x = 0$ or $x = 1$.

When $x =  - \dfrac{1}{2}$, from (1), we get ${y^2} = \dfrac{1}{4} + \dfrac{1}{2}$ or ${y^2} = \dfrac{3}{4}$, i.e., $y =  \pm \dfrac{{\sqrt 3 }}{2}$. 

Hence, the solutions of the given equation are

$0 + i0,1 + i0, - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}, - \dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}$

Example 4. If the imaginary part of $\dfrac{{2z + 1}}{{iz + 1}}$ is $- 2$, then show that the locus of the point representing $z$ in the argand plane is a straight line.

Ans: Given, the imaginary part of $\dfrac{{2z + 1}}{{iz + 1}}$ is $- 2$.

Let $z = x + iy$ and simplify the given equation then compare the real and imaginary parts.

Let $z = x + iy$. Then

$\dfrac{{2z + 1}}{{iz + 1}} = \dfrac{{2(x + iy) + 1}}{{i(x + iy) + 1}}$

 $\Rightarrow \dfrac{{(2x + 1) + i2y}}{{(1 - y) + ix}} = \dfrac{{\{ (2x + 1) + i2y\} }}{{\{ (1 - y) + ix\} }} \times \dfrac{{\{ (1 - y) - ix\} }}{{\{ (1 - y) - ix\} }}$

 $\Rightarrow \dfrac{{(2x + 1 - y) + i\left( {2y - 2{y^2} - 2{x^2} - x} \right)}}{{1 + {y^2} - 2y + {x^2}}}$ 

Thus $\operatorname{Im} \left( {\dfrac{{2z + 1}}{{iz + 1}}} \right) = \dfrac{{2y - 2{y^2} - 2{x^2} - x}}{{1 + {y^2} - 2y + {x^2}}}$

But $ \operatorname{Im} \left( {\dfrac{{2z + 1}}{{iz + 1}}} \right) =  - 2\quad$ (Given)

So $\dfrac{{2y - 2{y^2} - 2{x^2} - x}}{{1 + {y^2} - 2y + {x^2}}} =  - 2$

$\Rightarrow  2y - 2{y^2} - 2{x^2} - x =  - 2 - 2{y^2} + 4y - 2{x^2}$

i.e., $ x + 2y - 2 = 0$, which is the equation of a line.

Example 5. If $\left| {{z^2} - 1} \right| = |z{|^2} + 1$, then show that $z$ lies on an imaginary axis.

Ans: Given, $\left| {{z^2} - 1} \right| = |z{|^2} + 1$

Let $z = x + iy$ and simplify the given equation then compare the real and imaginary parts.

Let $z = x + iy.$ Then $\left| {{z^2} - 1} \right| = |z{|^2} + 1$

$\Rightarrow  \left| {{x^2} - {y^2} - 1 + i2xy} \right| = |x + iy{|^2} + 1$

$\Rightarrow  {\left( {{x^2} - {y^2} - 1} \right)^2} + 4{x^2}{y^2} = {\left( {{x^2} + {y^2} + 1} \right)^2}$

$\Rightarrow  4{x^2} = 0\quad$ i.e., $ x = 0$

Hence $z$ lies on $y$-axis.

Example 6. Let ${z_1}$ and ${z_2}$ be two complex numbers such that $\overline {{z_1}}  + i{\bar z_2} = 0$ and $arg\left( {{z_1}{z_2}} \right) = \pi$. Then find $arg\left( {{z_1}} \right)$.

Ans: Given, ${z_1}$ and ${z_2}$ be two complex numbers such that $\overline {{z_1}}  + i{\bar z_2} = 0$ and $\arg \left( {{z_1}{z_2}} \right) = \pi$.

Let $z = x + iy$ and simplify the given equation then compare the real and imaginary parts. Use $\arg \left( {{z_1}{z_2}} \right) = \arg \left( {{z_1}} \right) + \arg \left( {{z_2}} \right)$.

Given that ${\bar z_1} + i{\bar z_2} = 0$

$\Rightarrow  {z_1} = i{z_2}$, i.e., ${z_2} =  - i{z_1}$

Thus $\arg \left( {{z_1}{z_2}} \right) = \arg {z_1} + \arg \left( { - i{z_1}} \right) = \pi$

$\Rightarrow \quad$ arg $\left( { - iz_1^2} \right) = \pi$

$\Rightarrow  \arg ( - i) + \arg \left( {z_1^2} \right) = \pi$

$\Rightarrow  \arg ( - i) + 2\arg \left( {{z_1}} \right) = \pi$

$\Rightarrow  \dfrac{{ - \pi }}{2} + 2\arg \left( {{z_1}} \right) = \pi$

$\Rightarrow  \arg \left( {{z_1}} \right) = \dfrac{{3\pi }}{4}$

Example 7. Let ${z_1}$ and ${z_2}$ be two complex numbers such that $\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$. Then show that arg $\left( {{z_1}} \right) - arg\left( {{z_2}} \right) = 0$.

Ans: Given, ${z_1}$ and ${z_2}$ be two complex numbers such that $\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$

Let the complex number be $z = x + iy$ then it’s polar form is $z = r\left( {\cos \theta  + i\sin \theta } \right)$. Where $r = \sqrt {{x^2} + {y^2}}$ and $\arg (z) = \theta  = {\tan ^{ - 1}}\dfrac{y}{x}$.

Let ${z_1} = {r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)$ and ${z_2} = {r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right)$

where $ {r_1} = \left| {{z_1}} \right|,\arg \left( {{z_1}} \right) = {\theta _1},{r_2} = \left| {{z_2}} \right|,\arg \left( {{z_2}} \right) = {\theta _2}.$

We have, $ \left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$

 $= \left| {{r_1}\left( {\cos {\theta _1} + \cos {\theta _2}} \right) + {r_2}\left( {\cos {\theta _2} + \sin {\theta _2}} \right)} \right| = {r_1} + {r_2}$

  $= r_1^2 + r_2^2 + 2{r_1}{r_2}\cos \left( {{\theta _1} - {\theta _2}} \right) = {\left( {{r_1} + {r_2}} \right)^2}$

 $\Rightarrow \cos \left( {{\theta _1} - {\theta _2}} \right) = 1$

 $\Rightarrow {\theta _1} - {\theta _2}{\text{ i}}{\text{.e}}{\text{., }}\arg {z_1} = \arg {z_2}$

$\therefore \arg {z_1} - \arg {z_2} = 0$ 

Hence proved.

Example 8. If ${z_1},{z_2},{z_3}$ are complex numbers such that $\left| {{z_1}} \right| = \left| {{z_2}} \right| = \left| {{z_3}} \right| = \left| {\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_2}}} + \dfrac{1}{{{z_3}}}} \right| = 1$, then find the value of $\left| {{z_1} + {z_2} + {z_3}} \right|$.

Ans: Given, ${z_1},{z_2},{z_3}$ are complex numbers such that $\left| {{z_1}} \right| = \left| {{z_2}} \right| = \left| {{z_3}} \right| = \left| {\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_2}}} + \dfrac{1}{{{z_3}}}} \right| = 1$.

$\left| {{z_1}} \right| = \left| {{z_2}} \right| = \left| {{z_3}} \right| = 1$

$\Rightarrow {\left| {{z_1}} \right|^2} = {\left| {{z_2}} \right|^2} = {\left| {{z_3}} \right|^2} = 1$

$\Rightarrow  {z_1}{\bar z_1} = {z_2}{\bar z_2} = {z_3}{\bar z_3} = 1$

$\Rightarrow  {\bar z_1} = \dfrac{1}{{{z_1}}},{\bar z_2} = \dfrac{1}{{{z_2}}},{\bar z_3} = \dfrac{1}{{{z_3}}}$

Given that 

$\left| {\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_2}}} + \dfrac{1}{{{z_3}}}} \right| = 1$

$\Rightarrow\left| {{{\bar z}_1} + {{\bar z}_2} + {{\bar z}_3}} \right| = 1$, i.e.,$ \left| {{z_1} + {z_2} + {z_3}} \right| = 1$

$\Rightarrow  \left| {{z_1} + {z_2} + {z_3}} \right| = 1$

Example 9. If a complex number $z$ lies in the interior or on the boundary of a circle of radius 3 units and centre $( - 4,0)$, find the greatest and least values of $|z + 1|$.

Ans: Distance of the point representing $z$ from the centre of the circle is $|z - ( - 4 + i0)| = |z + 4|$

According to given condition $|z + 4| \leqslant 3$.

Now,

$|z + 1| = |z + 4 - 3| \leqslant |z + 4| + | - 3| \leqslant 3 + 3 = 6$

Therefore, greatest value of $|z + 1|$ is 6.

Since least value of the modulus of a complex number is zero, the least value of

$|z + 1| = 0$.

Example 10. Locate the points for which $3 < |z| < 4$

Ans:  $|z| < 4 \Rightarrow {x^2} + {y^2} < 16$ which is the interior of circle with centre at origin and

radius 4 units, and $|z| > 3 \Rightarrow {x^2} + {y^2} > 9$ which is exterior of circle with centre at origin

and radius 3 units. Hence $3 < |z| < 4$ is the portion between two circles ${x^2} + {y^2} = 9$ and ${x^2} + {y^2} = 16$.

Example 11. Find the value of $2{x^4} + 5{x^3} + 7{x^2} - x + 41$, when $x =  - 2 - \sqrt 3 i$.

Ans: Given, $x =  - 2 - \sqrt 3 i$

Put real numbers on one side and imaginary on the other in the given equation $x =  - 2 - \sqrt 3 i$ and square both sides. To obtain the complete part as real number.

  $x + 2 =  - \sqrt 3 i\left( {{\text{Squaring both the sides}}} \right)$

 $\Rightarrow {x^2} + 4 + 4x =  - 3$

 $\Rightarrow {x^2} + 4x + 7 = 0$ 

Therefore, On factorisation

  $2{x^4} + 5{x^3} + 7{x^2} - x + 41$

 $\Rightarrow \left( {{x^2} + 4x + 7} \right)\left( {2{x^2} - 3x + 5} \right) + 6$

 $\Rightarrow 0 \times \left( {2{x^2} - 3x + 5} \right) + 6 = 6.$ 

Example 12. Find the value of ${\text{P}}$ such that the difference of the roots of the equation ${x^2} - Px + 8 = 0$ is $2$.

Ans: Given, Difference of the roots of the equation ${x^2} - {\text{P}}x + 8 = 0$ is $2$.

A standard quadratic equation looks like this: $a{x^2} + bx + c = 0$. the sum of the roots is $- \dfrac{b}{a}$ and the product of the roots is $\dfrac{c}{a}$. 

Let $\alpha ,\beta$ be the roots of the equation ${x^2} - Px + 8 = 0$

${{\text{ Therefore }}} {\alpha  + \beta  = P{\text{ and }}\alpha  \cdot \beta  = 8.}$ 

${{\text{ Now }}} {\alpha  - \beta  =  \pm \sqrt {{{(\alpha  + \beta )}^2} - 4\alpha \beta } }$ 

Therefore,

$ 2 =  \pm \sqrt {{P^2} - 32}$

$\Rightarrow{P^2} - 32 = 4$, i.e., $P =  \pm 6$

Example 13. Find the value of $a$ such that the sum of the squares of the roots of the equation ${x^2} - (a - 2)x - (a + 1) = 0$ is least.

Ans: Given, the squares of the roots of the equation ${x^2} - (a - 2)x - (a + 1) = 0$ is least.

A standard quadratic equation looks like this: $a{x^2} + bx + c = 0$. the sum of the roots is $- \dfrac{b}{a}$ and the product of the roots is $\dfrac{c}{a}$.

Let $\alpha ,\beta$ be the roots of the equation 

${\text{Therefore,}}$

  $\alpha  + \beta  = a - 2{\text{ and }}\alpha \beta  =  - (a + 1){\text{ }}$

${\text{Now,}}$

${\alpha ^2} + {\beta ^2} = {(\alpha  + \beta )^2} - 2\alpha \beta$ 

$= {(a - 2)^2} + 2(a + 1)$

$= {(a - 1)^2} + 5$ 

Therefore,

${\alpha ^2} + {\beta ^2}$ will be minimum if ${(a - 1)^2} = 0$, i.e., $a = 1$.

Example 14. Find the value of $k$ if for the complex numbers ${z_1}$ and ${z_2}$,

${\left| {1 - {{\bar z}_1}{z_2}} \right|^2} - {\left| {{z_1} - {z_2}} \right|^2} = k\left( {1 - {{\left| {{z_1}} \right|}^2}} \right)\left( {1 - {{\left| {{z_2}} \right|}^2}} \right)$

Ans: L.H.S.$= {\left| {1 - {{\bar z}_1}{z_2}} \right|^2} - {\left| {{z_1} - {z_2}} \right|^2}$

 $\Rightarrow \left( {1 - {{\bar z}_1}{z_2}} \right)\left( {\overline {1 - {{\bar z}_1}{z_2}} } \right)\left( {{z_1} - {z_2}} \right)\left( {\overline {{z_1} - {z_2}} } \right)$

 $\Rightarrow \left( {1 - {{\bar z}_1}{z_2}} \right)\left( {1 - {z_1}{{\bar z}_2}} \right)\left( {{z_1} - {z_2}} \right)\left( {{{\bar z}_1} - {{\bar z}_2}} \right)$

 $\Rightarrow 1 + {z_1}{{\bar z}_1}{z_2}{{\bar z}_2} - {z_1}{{\bar z}_1} - {z_2}{{\bar z}_2}$

 $\Rightarrow 1 + {\left| {{z_1}} \right|^2}.{\left| {{z_2}} \right|^2} - {\left| {{z_1}} \right|^2} - {\left| {{z_2}} \right|^2}$

 $\Rightarrow \left( {1 - {{\left| {{z_1}} \right|}^2}} \right)\left( {1 - {{\left| {{z_2}} \right|}^2}} \right)$ 

R.H.S.  $k\left( {1 - {{\left| {{z_1}} \right|}^2}} \right)\left( {1 - {{\left| {{z_2}} \right|}^2}} \right)$

 $\Rightarrow k = 1$ 

Hence, equating LHS and RHS, we get $k = 1$.

Example 15. If ${z_1}$ and ${z_2}$ both satisfy $z + \bar z = 2|z - 1|$ arg $\left( {{z_1} - {z_2}} \right) = \dfrac{\pi }{4}$, then find $Im\left( {{z_1} + {z_2}} \right)$.

Ans: Given, ${z_1}$ and ${z_2}$ both satisfy $z + \bar z = 2|z - 1|$ arg $\left( {{z_1} - {z_2}} \right) = \dfrac{\pi }{4}$.

Let the complex number be $z = x + iy$ then it’s polar form is $z = r\left( {\cos \theta  + i\sin \theta } \right)$. Where $r = \sqrt {{x^2} + {y^2}}$ and $\arg (z) = \theta  = {\tan ^{ - 1}}\dfrac{y}{x}$.

Let $z = x + iy,{z_1} = {x_1} + i{y_1}$ and ${z_2} = {x_2} + i{y_2}$

Then $ z + \bar z = 2|z - 1|$

$\Rightarrow(x + iy) + (x - iy) = 2|x - 1 + iy|$

$\Rightarrow  2x = 1 + {y^2}.....\left( 1 \right)$

Since ${z_1}$ and ${z_2}$ both satisfy (1), we have

$2{x_1} = 1 + y_1^2 \ldots$ and $2{x_2} = 1 + y_2^2$

$\Rightarrow  2\left( {{x_1} - {x_2}} \right) = \left( {{y_1} + {y_2}} \right)\left( {{y_1} - {y_2}} \right)$

$\Rightarrow  2 = \left( {{y_1} + {y_2}} \right)\left( {\dfrac{{{y_1} - {y_2}}}{{{x_1} - {x_2}}}} \right).....\left( 2 \right)$

Again 

${z_1} - {z_2} = \left( {{x_1} - {x_2}} \right) + i\left( {{y_1} - {y_2}} \right)$

Therefore, 

$\tan \theta  = \dfrac{{{y_1} - {y_2}}}{{{x_1} - {x_2}}}$, where $\theta  = \arg \left( {{z_1} - {z_2}} \right)$

$\Rightarrow\tan \dfrac{\pi }{4} = \dfrac{{{y_1} - {y_2}}}{{{x_1} - {x_2}}}\left( {{\text{Since }}\theta  = \dfrac{\pi }{4}} \right)$

i.e., $ 1 = \dfrac{{{y_1} - {y_2}}}{{{x_1} - {x_2}}}$

From (2), we get

$2 = {y_1} + {y_2}$, i.e., $\operatorname{Im} \left( {{z_1} + {z_2}} \right) = 2$

Example 16. Fill in the blanks:

(i) The real value of ' $a$ ' for which $3{i^3} - 2a{i^2} + (1 - a)i + 5$ is real is _______.

Ans: $3{i^3} - 2a{i^2} + (1 - a)i + 5$ =  - 3i + 2a + 5 + (1 - a)i

$= 2a + 5 + ( - a - 2)i$, which is real if $- a - 2 = 0$ i.e. $a =  - 2$.

(ii) If $|z| = 2$ and $arg(z) = \dfrac{\pi }{4}$, then $z =$_______.

Ans: $z = |z|\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)$

 $\Rightarrow 2\left( {\dfrac{1}{{\sqrt 2 }} + i\dfrac{1}{{\sqrt 2 }}} \right) = \sqrt 2 (1 + i)$ 

(iii) The locus of $z$ satisfying arg $(z) = \dfrac{\pi }{3}$ is _______.

Ans: Let $z = x + iy.$ Then its polar form is $z = r(\cos \theta  + i\sin \theta )$, where $\tan \theta  = \dfrac{y}{x}$ and

$\theta$ is arg $(z)$. Given that $\theta  = \dfrac{\pi }{3}$. Thus. $\tan \dfrac{\pi }{3} = \dfrac{y}{x} \Rightarrow y = \sqrt 3 x$, where $x > 0,y > 0$.

Hence, locus of $z$ is the part of $y = \sqrt 3 x$ in the first quadrant except origin.

(iv) The value of ${( - \sqrt { - 1} )^{4n - 3}}$, where $n\in N$, is ________.

Ans: Here ${( - \sqrt { - 1} )^{4n - 3}} = {( - i)^{4n - 3}} = {( - i)^{4n}}{( - i)^{ - 3}} = \dfrac{1}{{{{( - i)}^3}}}$

$\dfrac{1}{{ - {i^3}}} = \dfrac{1}{i}$

 $\Rightarrow \dfrac{i}{{{i^2}}} =  - i$ 

(v) The conjugate of the complex number $\dfrac{{1 - i}}{{1 + i}}$ is _________.

Ans: $\dfrac{{1 - i}}{{1 + i}} = \dfrac{{1 - i}}{{1 + i}} \times \dfrac{{1 - i}}{{1 - i}} = \dfrac{{1 + {i^2} - 2i}}{{1 - {i^2}}} = \dfrac{{1 - 1 - 2i}}{{1 + 1}} =  - i$

Hence, conjugate of $\dfrac{{1 - i}}{{1 + i}}$ is $i$.

(vi) If a complex number lies in the third quadrant, then its conjugate lies in the ________.

Ans: Conjugate of a complex number is the image of the complex number about the $x$-axis. Therefore, if a number lies in the third quadrant, then its image lies in the second quadrant.

(vii) If $(2 + i)(2 + 2i)(2 + 3i) \ldots (2 + ni) = x + iy$, then $5.8.13 \ldots \left( {4 + {n^2}} \right) =$________.

Ans: Given that,

$(2 + i)(2 + 2i)(2 + 3i) \ldots (2 + ni) = x + iy$              …..(1)

$\Rightarrow  (\overline {2 + i} )(\overline {2 + 2i} )(\overline {2 + 3i} ) \ldots (\overline {(2 + ni} ) = (\overline {x + iy} ) = (x - iy)$

i.e., $ (2 - i)(2 - 2i)(2 - 3i) \ldots (2 - ni) = x - iy$        …..(2)

Multiplying (1) and (2), we get $5.8.13...\left( {4 + {n^2}} \right) = {x^2} + {y^2}$.

Example 17. State true or false for the following:

(i) Multiplication of a non-zero complex number by $i$ rotates it through a right angle in the anti-clockwise direction.

Ans: True. Let $z = 2 + 3i$ be a complex number represented by OP. Then $iz =  - 3 + 2i$, represented by ${\text{OQ}}$, where if OP is rotated in the anticlockwise direction through a right angle, it coincides with OQ.

(ii) The complex number $cos\theta  + isin\theta$ can be zero for some $\theta$.

Ans: False. Because $\cos \theta  + i\sin \theta  = 0 \Rightarrow \cos \theta  = 0$ and $\sin \theta  = 0.$ But there is no

value of $\theta$ for which $\cos \theta$ and $\sin \theta$ both are zero.

(iii) If a complex number coincides with its conjugate, then the number must lie on imaginary axis.

Ans: False, because $x + iy = x - iy \Rightarrow y = 0 \Rightarrow$ number lies on $x$-axis.

(iv) The argument of the complex number $z = (1 + i\sqrt 3 )(1 + i)(cos\theta  + isin\theta )$ is $\dfrac{{7\pi }}{{12}} + \theta$

Ans: True, $\arg (z) = \arg (1 + i\sqrt 3 ) + \arg (1 + i) + \arg (\cos \theta  + i\sin \theta )$

$\dfrac{\pi }{3} + \dfrac{\pi }{4} + \theta  = \dfrac{{7\pi }}{{12}} + \theta$

(v) The points representing the complex number $z$ for which $|z + 1| < |z - 1|$ lies in the interior of a circle.

Ans: False, because $|x + iy + 1| < |x + iy - 1|$

$\Rightarrow{(x + 1)^2} + {y^2} < {(x - 1)^2} + {y^2}$ which gives $4x < 0$

(vi) complex numbers ${z_1},{z_2}$ and ${z_3}$ are in A.P., then they lie on a circle in the complex plane. (vi) If three 

Ans: False, because if ${z_1},{z_2}$ and ${{\text{z}}_3}$ are in A.P., then

${z_2} = \dfrac{{{z_1} + {z_3}}}{2}$

$\Rightarrow {z_2}$ is the midpoint of ${z_1}$ and ${z_3}$, which implies that the points ${z_1},{z_2},{z_3}$ are collinear.

(vii) If $n$ is a positive integer, then the value of ${i^n} + {(i)^{n + 1}} + {(i)^{n + 2}} + {(i)^{n + 3}}$ is 0.

Ans: True, because ${i^n} + {(i)^{n + 1}} + {(i)^{n + 2}} + {(i)^{n + 3}}$

$= {i^n}\left( {1 + i + {i^2} + {i^3}} \right) = {i^n}(1 + i - 1 - i)$

 $\Rightarrow {i^n}(0) = 0$ 

Example 18. Match the statements of Column A and B.

Ans:  (a) $\Leftrightarrow$ (ii), because $1 + {i^2} + {i^4} + {i^6} +  \ldots  + {i^{20}}$

$= 1 - 1 + 1 - 1 +  \ldots  + 1 = 1$ (which is purely a real complex number)

(b) $\Leftrightarrow$ (i), because ${i^{ - 1007}} = \dfrac{1}{{{{(i)}^{1097}}}} = \dfrac{1}{{{i^{4 \times 274 + 1}}}} = \dfrac{1}{{{{\left\{ {{{(i)}^4}} \right\}}^{274}}(i)}} = \dfrac{1}{i} = \dfrac{i}{{{i^2}}} =  - i$

which is purely imaginary complex number.

(c) $\Leftrightarrow$ (iv), conjugate of $1 + i$ is $1 - i$, which is represented by the point $(1, - 1)$ in the fourth quadrant.

(d) $\Leftrightarrow$ (iii), because $\dfrac{{1 + 2i}}{{1 - i}} = \dfrac{{1 + 2i}}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}} = \dfrac{{ - 1 + 3i}}{2} =  - \dfrac{1}{2} + \dfrac{3}{2}i$, which is

represented by the point $\left( { - \dfrac{1}{2},\dfrac{3}{2}} \right)$ in the second quadrant.

(e) $\Leftrightarrow ({\text{vi}})$, If ${b^2} - 4ac < 0 = {\text{D}} < 0$, i.e., square root of ${\text{D}}$ is a imaginary

number, therefore, roots are $x = \dfrac{{ - b \pm {\text{ Imaginary Number }}}}{{2a}}$, i.e., roots are in conjugate pairs.

(f) $\Leftrightarrow$ (v), Consider the equation ${x^2} - (5 + \sqrt 2 )x + 5\sqrt 2  = 0$, where $a = 1$, $b =  - (5 + \sqrt 2 ),c = 5\sqrt 2$, clearly $a,b,c \in {\text{R}}$

Now ${\text{D}} = {b^2} - 4ac = {\{  - (5 + \sqrt 2 )\} ^2} - 4 \cdot 1.5\sqrt 2  = {(5 - \sqrt 2 )^2}$.

therefore$x = \dfrac{{5 + \sqrt 2  \pm 5 - \sqrt 2 }}{2} = 5,\sqrt 2$ which do not form a conjugate pair.

Example 19. What is the value of $\dfrac{{{i^{4n + 1}} - {i^{4n - 1}}}}{2}$ ?

Ans: ${\text{ }}\dfrac{{{i^{4n + 1}} - {i^{4n - 1}}}}{2} = \dfrac{{{i^{4n}}i - {i^{4n}}{i^{ - i}}}}{2}$

 $\Rightarrow \dfrac{{i - \dfrac{1}{i}}}{2} = \dfrac{{{i^2} - 1}}{{2i}}$ 

$\Rightarrow \dfrac{{ - 2}}{{2i}} = i$

Example 20. What is the smallest positive integer $n$, for which ${(1 + i)^{2n}} = {(1 - i)^{2n}}$ ?

Ans: $n = 2$, because,

${(1 + i)^{2n}} = {(1 - i)^{2n}}$

 $\Rightarrow {\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^{2n}} = 1$

 $\Rightarrow {\left( i \right)^{2n}} = 1{\text{which is possible if }}n = 2\left( {\because {i^4} = 1} \right)$ 

Example 21. What is the reciprocal of $3 + \sqrt 7 i$

Ans: Given, $3 + \sqrt 7 i$

Reciprocal of $z = \dfrac{{\bar z}}{{|z{|^2}}}$

Therefore, reciprocal of $3 + \sqrt 7 i = \dfrac{{3 - \sqrt 7 i}}{{16}} = \dfrac{3}{{16}} - \dfrac{{\sqrt 7 i}}{{16}}$

Example 22. If ${z_1} = \sqrt 3  + i\sqrt 3$ and ${z_2} = \sqrt 3  + i$, then find the quadrant in which $\left( {\dfrac{{{z_1}}}{{{z_2}}}} \right)$ lies.

Ans: $\dfrac{{{z_1}}}{{{z_2}}} = \dfrac{{\sqrt 3  + i\sqrt 3 }}{{\sqrt 3  + i}} = \left( {\dfrac{{3 + \sqrt 3 }}{4}} \right) + \left( {\dfrac{{3 - \sqrt 3 }}{4}} \right)i$

which is represented by a point in the first quadrant.

Example 23. What is the conjugate of $\dfrac{{\sqrt {5 + 12i}  + \sqrt {5 - 12i} }}{{\sqrt {5 + 12i}  - \sqrt {5 - 12i} }}$ ?

Ans: Let,

$z = \dfrac{{\sqrt {5 + 12i}  + \sqrt {5 - 12i} }}{{\sqrt {5 + 12i}  - \sqrt {5 - 12i} }} \times \dfrac{{\sqrt {5 + 12i}  + \sqrt {5 - 12i} }}{{\sqrt {5 + 12i}  + \sqrt {5 - 12i} }}$

 $\Rightarrow \dfrac{{5 + 12i + 5 - 12i + 2\sqrt {25 + 144} }}{{5 + 12i - 5 + 12i}} = \dfrac{3}{{2i}}$

 $\Rightarrow \dfrac{{3i}}{{ - 2}} = 0 - \dfrac{3}{2}i$ 

Therefore, the conjugate of $z = 0 + \dfrac{3}{2}i$

Example 24. What is the principal value of amplitude of $1 - i$ ?

Ans: Let $\theta$ be the principal value of amplitude of $1 - i.$

 Since  $\tan \theta  =  - 1$

 $\Rightarrow \tan \theta  = \tan \left( { - \dfrac{\pi }{4}} \right)$

 $\Rightarrow \theta  =  - \dfrac{\pi }{4}$ 

Example 25. What is the polar form of the complex number ${\left( {{i^{25}}} \right)^3}?$

Ans: $z = {\left( {{i^{25}}} \right)^3}$

 $\Rightarrow {(i)^{75}} = {i^{4 \times 18 + 3}}$

 $\Rightarrow {\left( {{i^4}} \right)^{18}}{(i)^3}$ 

$= {i^3} =  - i = 0 - i$

Polar form of $z = r(\cos \theta  + i\sin \theta )$

$= 1\left\{ {\cos \left( { - \dfrac{\pi }{2}} \right) + i\sin \left( { - \dfrac{\pi }{2}} \right)} \right\}$

 $\Rightarrow \cos \dfrac{\pi }{2} - i\sin \dfrac{\pi }{2}$ 

Example 26. What is the locus of $z$, if amplitude of $z - 2 - 3i$ is $\dfrac{\pi }{4}?$

Ans: Let $z = x + iy.$

Then $z - 2 - 3i = (x - 2) + i(y - 3)$

Let $\theta$ be the amplitude of $z - 2 - 3i$.

Then $\tan \theta  = \dfrac{{y - 3}}{{x - 2}}$

$\Rightarrow  \tan \dfrac{\pi }{4} = \dfrac{{y - 3}}{{x - 2}}\left( {\operatorname{since} ,\theta  = \dfrac{\pi }{4}} \right)$

$\Rightarrow 1 = \dfrac{{y - 3}}{{x - 2}}i.e.,x - y + 1 = 0$

Example 27. If $1 - i$, is a root of the equation ${x^2} + ax + b = 0$, where $a,b\in R$, then find the values of $a$ and $b$.

Ans: Sum of roots 

$\dfrac{{ - a}}{1} = (1 - i) + (1 + i)$

 $\Rightarrow a =  - 2$ 

 (since non real complex roots occur in conjugate pairs) 

Product of roots, 

$\dfrac{b}{1} = (1 - i)(1 + i)$

 $\Rightarrow b = 2$ 

Example 28. $1 + {i^2} + {i^4} + {i^6} +  \ldots  + {i^{2n}}$ is

(A) positive

(B) negative

(C) 0

(D) can not be evaluated

Ans: Option D is correct answer

 $1 + {i^2} + {i^4} + {i^6} +  \ldots  + {i^{2n}}$

The value of $i = \sqrt { - 1} {\text{, }}{i^2} =  - 1,{i^3} =  - i{\text{ and }}{i^4} = 1$

$1 + {i^2} + {i^4} + {i^6} +  \ldots  + {i^{2n}} = 1 - 1 + 1 - 1 +  \ldots {( - 1)^n}$

which can not be evaluated unless $n$ is known.

Example 29. If the complex number $z = x + iy$ satisfies the condition $|z + 1| = 1$, then $z$ lies on

(A) $x$-axis

(B) circle with centre $(1,0)$ and radius 1

(C) circle with centre $( - 1,0)$ and radius 1

(D) $y$-axis

Ans: Option C is correct answer

${\text{Let,}}$

$z = x + iy{\text{  then,}}$

  $|z + 1| = 1$

 $\Rightarrow |(x + 1) + iy| = 1$ 

$\Rightarrow  {(x + 1)^2} + {y^2} = 1$

which is a circle with center $( - 1,0)$ and radius 1.

Example 30. The area of the triangle on the complex plane formed by the complex numbers $z, - iz$ and $z + iz$ is:

(A) $|z{|^2}$

(B) $|\bar z{|^2}$

(C) $\dfrac{{|z{|^2}}}{2}$

(D) none of these

Ans: Option C is correct answer

 Let $z = x + iy$.

Then $- iz = y - ix$.

Therefore, $z + iz = (x - y) + i(x + y)$

Required area of the triangle $= \dfrac{1}{2}\left( {{x^2} + {y^2}} \right) = \dfrac{{|z{|^2}}}{2}$

Example 31. The equation $|z + 1 - i| = |z - 1 + i|$ represents a

(A) straight line

(B) circle

(C) parabola

(D) hyperbola

Ans: Option A is correct answer

 $|z + 1 - i| = |z - 1 + i|$

$\Rightarrow  |z - ( - 1 + i)| = |z - (1 - i)|$

$\Rightarrow  {\text{PA}} = {\text{PB}}$, where ${\text{A}}$ denotes the point $( - 1,1),{\text{B}}$ denotes the point $(1, - 1)$ and ${\text{P}}$ denotes the point $(x,y)$

$\Rightarrow  z$ lies on the perpendicular bisector of the line joining ${\text{A}}$ and ${\text{B}}$ and the perpendicular bisector is a straight line. 

Example 32. Number of solutions of the equation ${z^2} + |z{|^2} = 0$ is

(A) 1

(B) 2

(C) 3

(D) infinitely many

Ans: Option D is correct answer

 ${z^2} + |z{|^2} = 0,z \ne 0$

$\Rightarrow  {x^2} - {y^2} + i2xy + {x^2} + {y^2} = 0$

$\Rightarrow  2{x^2} + i2xy = 0 \Rightarrow 2x(x + iy) = 0$

$\Rightarrow  x = 0$ or $x + iy = 0$ (not possible)

Therefore, $x = 0$ and $z \ne 0$

So $y$ can have any real value. Hence infinitely many solutions.

Example 33. The amplitude of $sin\dfrac{\pi }{5} + i\left( {1 - cos\dfrac{\pi }{5}} \right)$ is

(A) $\dfrac{{2\pi }}{5}$

(B) $\dfrac{\pi }{5}$

(C) $\dfrac{\pi }{{15}}$

(D) $\dfrac{\pi }{{10}}$

Ans: Option D is correct answer

Here $r\cos \theta  = \sin \left( {\dfrac{\pi }{5}} \right)$ and $r\sin \theta  = 1 - \cos \dfrac{\pi }{5}$

Therefore, $\tan \theta  = \dfrac{{1 - \cos \dfrac{\pi }{5}}}{{\sin \dfrac{\pi }{5}}} = \dfrac{{2{{\sin }^2}\left( {\dfrac{\pi }{{10}}} \right)}}{{2\sin \left( {\dfrac{\pi }{{10}}} \right) \cdot \cos \left( {\dfrac{\pi }{{10}}} \right)}}$

$\Rightarrow  \tan \theta  = \tan \left( {\dfrac{\pi }{{10}}} \right)$ i.e., $\theta  = \dfrac{\pi }{{10}}$

Exercise

Short Answer Type Questions

1. For a positive integer $n$, find the value of ${\left( {1 - i} \right)^n}{\left( {1 - \dfrac{1}{i}} \right)^n}$

Ans: Given, ${\left( {1 - i} \right)^n}{\left( {1 - \dfrac{1}{i}} \right)^n}$

The value of $i = \sqrt { - 1} {\text{  and }}{i^2} =  - 1$

Given expression,

$= {(1 - i)^n}{\left( {1 - \dfrac{1}{i}} \right)^n}$

 $\Rightarrow {(1 - i)^n}{\left( {\dfrac{{i - 1}}{i}} \right)^n}$

 $\Rightarrow {(1 - i)^n}{(i - 1)^n} \cdot {i^{ - n}} = {(1 - i)^n}{(1 - i)^n}{( - 1)^n} \cdot {i^{ - n}}$

 $\Rightarrow {\left[ {{{(1 - i)}^2}} \right]^n}{( - 1)^n} \cdot {i^{ - n}} = {\left( {1 + {i^2} - 2i} \right)^n}{( - 1)^n}{i^{ - n}}\left[ {\because {i^2} =  - 1} \right]$

 $\Rightarrow {(1 - 1 - 2i)^n}{( - 1)^n}{i^{ - n}} = {( - 2)^n} \cdot {i^n}{( - 1)^n}{i^{ - n}}$

 $\Rightarrow {( - 1)^{2n}} \cdot {2^n} = {2^n}$

 $\Rightarrow 1 \cdot {2^n} = {2^n}$ 

2. Evaluate $\sum\limits_{n = 1}^{13} {\left( {{i^n} + {i^{n + 1}}} \right)}$ ,where $n\in N$.

Ans: Given, $\sum\limits_{n = 1}^{13} {\left( {{i^n} + {i^{n + 1}}} \right)}$ ,where $n \in N$.

The value of $i = \sqrt { - 1} {\text{, }}{i^2} =  - 1,{i^3} =  - i{\text{ and }}{i^4} = 1$

  $\sum\limits_{n = 1}^{13} {\left( {{i^n} + {i^{n + 1}}} \right)} ,n \in N = \sum\limits_{h = 1}^{13} {{i^n}} (1 + i)$

 $\Rightarrow (1 + i)\left[ {i + {i^2} + {i^3} + {i^4} + {i^5} + {i^6} + {i^7} + {i^8} + {i^9} + {i^{10}} + {i^{11}} + {i^{12}} + {i^{13}}} \right]$

 $\Rightarrow (1 + i)\left[ {{i^{13}}} \right]\left( {\because {i^n} + {i^{n + 1}} + {i^{n + 2}} + {i^{n + 3}} = 0,{\text{ where }}n \in N{\text{ i}}{\text{.e}}{\text{., }}\sum\limits_{n = 1}^{12} {{i^n}}  = 0} \right)$

 $\Rightarrow (1 + i)i = \left( {{i^2} + i} \right)$

 $\because i = \sqrt { - 1}$

$\therefore {i^2} =  - 1$ 

 $\Rightarrow i - 1$ 

3. If ${\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^3} - {\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^3} = x + iy$, then find $\left( {x,y} \right)$.

Ans: Given, ${\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^3} - {\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^3} = x + iy$

If two complex numbers ${{\text{z}}_1} = {x_1} + i{y_1}$ and ${{\text{z}}_2} = {x_2} + i{y_2}$ are equal

  i.e.,  $ {z_1} = {z_2} \Rightarrow {x_1} + i{y_1} = {x_2} + i{y_2}$ then ${x_1} = {x_2}$ and ${y_1} = {y_2}$.

Given that, ${\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^3} - {\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^3} = x + \dot y.....\left( i \right)$

$\therefore  {\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^3} = \dfrac{{1 + {i^3} + 3i(1 + i)}}{{1 - {i^3} - 3i(1 - i)}} = \dfrac{{1 - i + 3i + 3{i^2}}}{{1 + i - 3i + 3{i^2}}}$

$= \dfrac{{2i - 2}}{{ - 2i - 2}} = \dfrac{{i - 1}}{{ - i - 1}} = \dfrac{{1 - i}}{{1 + i}}$

$= \dfrac{{(1 - i)}}{{(1 + i)}}\dfrac{{(1 - i)}}{{(1 - i)}} = \dfrac{{1 + {i^2} - 2i}}{{1 + 1}} = \dfrac{{1 - 1 - 2i}}{2}$

$\Rightarrow{\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^3} =  - i.....\left( {ii} \right)$

Similarly,    ${\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^3} = \dfrac{{ - 1}}{i} = \dfrac{{{i^2}}}{i} = i.....\left( {iii} \right)$

Using Eqs. (ii) and (iii) in Eq. (i), we get

   $- i - i = x + iy$

 $\Rightarrow  - 2i = x + iy$ 

On comparing real and imaginary part of complex number, we get

$x = 0$ and $y =  - 2$

So,

$(x,y) = (0, - 2)$

4. If $\dfrac{{{{\left( {1 + i} \right)}^2}}}{{2 - i}} = x + iy,$ then find the value of $x + y$. 

Ans: Given, $\dfrac{{{{\left( {1 + i} \right)}^2}}}{{2 - i}} = x + iy,$

If two complex numbers ${{\text{z}}_1} = {x_1} + i{y_1}$ and ${{\text{z}}_2} = {x_2} + i{y_2}$ are equal

 i.e.,  $ {z_1} = {z_2} \Rightarrow {x_1} + i{y_1} = {x_2} + i{y_2}$ then ${x_1} = {x_2}$ and ${y_1} = {y_2}$.

${\text{ Given that,}}$

$\dfrac{{{{(1 + i)}^2}}}{{2 - i}} = x + iy$

 $\Rightarrow \dfrac{{\left( {1 + {i^2} + 2i} \right)}}{{2 - i}} = x + iy$

 $\Rightarrow \dfrac{{2i}}{{2 - i}} = x + iy$

 $\Rightarrow \dfrac{{2i(2 + i)}}{{(2 - i)(2 + i)}} = x + iy$

 $\Rightarrow \dfrac{{4i + 2{i^2}}}{{4 - {i^2}}} = x + iy$

 $\Rightarrow  \dfrac{{4i - 2}}{{4 + 1}} = x + iy$

 $\Rightarrow \dfrac{{ - 2}}{5} + \dfrac{{4i}}{5} = x + iy$ 

On comparing both sides, we get $x =  - \dfrac{2}{5}{\text{and }}y = \dfrac{4}{5}$

$\Rightarrow  x + y = \dfrac{{ - 2}}{5} + \dfrac{4}{5} = \dfrac{2}{5}$

5. If ${\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^{100}} = a + ib,$ then find $\left( {a,b} \right)$.

Ans: Given, ${\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^{100}} = a + ib,$

The value of $i = \sqrt { - 1} {\text{, }}{i^2} =  - 1,{i^3} =  - i{\text{ and }}{i^4} = 1$

 If two complex numbers ${{\text{z}}_1} = {x_1} + i{y_1}$ and ${{\text{z}}_2} = {x_2} + i{y_2}$ are equal

 i.e.,  $ {z_1} = {z_2} \Rightarrow {x_1} + i{y_1} = {x_2} + i{y_2}$ then ${x_1} = {x_2}$ and ${y_1} = {y_2}$.

Given that, ${\left( {\dfrac{{1 - i}}{{1 + i}}} \right)^{100}} = a + ib$

 $\Rightarrow  {\left[ {\dfrac{{(1 - i)}}{{(1 + i)}} \cdot \dfrac{{(1 - i)}}{{(1 - i)}}} \right]^{100}} = a + ib$

 $\Rightarrow {\left( {\dfrac{{1 + {i^2} - 2i}}{{1 - {i^2}}}} \right)^{100}} = a + ib$ 

$\Rightarrow  {\left( {\dfrac{{ - 2i}}{2}} \right)^{100}} = a + ib \left[ {\because {i^2} =  - 1} \right]$

 $\Rightarrow {\left( {{i^4}} \right)^{25}} = a + ib$

 $\Rightarrow 1 = a + ib$ 

Then, $a = 1$ and $b = 0 \left[ {\because {i^4} = 1} \right]$

$\therefore (a,b) = (1,0)$

6: If $a = cos\theta  + isin\theta ,$ find the value of $\dfrac{{1 + a}}{{1 - a}}$.

Ans: Given, $a = \cos \theta  + i\sin \theta ,$

The value of $i = \sqrt { - 1} {\text{, }}{i^2} =  - 1,{i^3} =  - i{\text{ and }}{i^4} = 1$

To solve the above problem use the trigonometric formula, $\cos \theta  = 2{\cos ^2}\dfrac{\theta }{2}- 1 = 1 - 2{\sin ^2}\dfrac{\theta }{2}$ and $\sin \theta  = 2\sin \dfrac{\theta }{2} \cdot \cos \dfrac{\theta }{2}$

Given that,$a = \cos \theta  + i\sin \theta$

$\therefore \dfrac{{1 + a}}{{1 - a}} = \dfrac{{1 + \cos \theta  + i\sin \theta }}{{1 - \cos \theta  - i\sin \theta }}$

 $\Rightarrow \dfrac{{1 + 2{{\cos }^2}\dfrac{\theta }{2} - 1 + 2i\sin \dfrac{\theta }{2} \cdot \cos \dfrac{\theta }{2}}}{{1 - 1 + 2{{\sin }^2}\dfrac{\theta }{2} - 2i\sin \dfrac{\theta }{2} \cdot \cos \dfrac{\theta }{2}}}$

 $\Rightarrow \dfrac{{2\cos \dfrac{\theta }{2}\left( {\cos \dfrac{\theta }{2} + i\sin \dfrac{\theta }{2}} \right)}}{{2\sin \dfrac{\theta }{2}\left( {\sin \dfrac{\theta }{2} - i\cos \dfrac{\theta }{2}} \right)}}$ 

 $\Rightarrow  - \dfrac{{2\cos \dfrac{\theta }{2}\left( {\cos \dfrac{\theta }{2} + i\sin \dfrac{\theta }{2}} \right)}}{{2i\sin \dfrac{\theta }{2}\left( {\cos \dfrac{\theta }{2} + i\sin \dfrac{\theta }{2}} \right)}} =  - \dfrac{1}{i}\cot \dfrac{\theta }{2}$

 $\Rightarrow \dfrac{{ + {i^2}}}{i}\cot \dfrac{\theta }{2} = i\cot \dfrac{\theta }{2} \left[ {\because \dfrac{{ - 1}}{i} = \dfrac{{{i^2}}}{i}} \right]$ 

7. If $\left( {1 + i} \right)z = \left( {1 - i} \right)\bar z,$ then show that $z =  - i\bar z$.

Ans: Given, $\left( {1 + i} \right)z = \left( {1 - i} \right)\bar z,$

The value of $i = \sqrt { - 1} {\text{, }}{i^2} =  - 1,{i^3} =  - i{\text{ and }}{i^4} = 1$

${\text{We have,}}$

  $(1 + i)z = (1 - i)\bar z$

 $\Rightarrow \dfrac{z}{{\bar z}} = \dfrac{{(1 - i)}}{{(1 + i)}}$

 $\Rightarrow  \dfrac{z}{{\bar z}} = \dfrac{{(1 - i)}}{{(1 + i)}}\dfrac{{(1 - i)}}{{(1 - i)}}$

 $\Rightarrow \dfrac{z}{{\bar z}} = \dfrac{{1 + {i^2} - 2i}}{{1 - {i^2}}} \left[ {\because {i^2} =  - 1} \right]$

 $\Rightarrow  \dfrac{z}{{\bar z}} = \dfrac{{1 - 1 - 2i}}{2}$

 $\Rightarrow \dfrac{z}{{\bar z}} =  - i$

$\therefore z = i\bar z$ 

Hence proved.

8. If $z = x + iy$, then show that $z\bar z + 2(z + \bar z) + b = 0$, where $b\in R$, represents a circle.

Ans: Given, $z = x + iy$

If $z = x + iy$ it’s conjugate is $\bar z = x - iy$.

If $z = x + iy$, then show that $z\bar z + 2(z + \bar z) + b = 0$, where $b \in R$, represents

a circle.

Given that, 

$z = x + iy$

Then, $\bar z = x - iy$

Now,

$z\bar z + 2(z + \bar z) + b = 0$

$\Rightarrow  (x + iy)(x - iy) + 2(x + iy + x - iy) + b = 0$

$\Rightarrow {x^2} + {y^2} + 4x + b = 0$, which is the equation of a circle.

9. If the real part of $\dfrac{{\bar z + 2}}{{\bar z - 1}}$ is 4, then show that the locus of the point representing $z$ in the complex plane is a circle.

Ans: Given, Real part of $\dfrac{{\bar z + 2}}{{\bar z - 1}}$ is 4

Evaluate the value of $\dfrac{{\bar z + 2}}{{\bar z - 1}}$ and if it satisfies the equation of the circle the locus of the point representing $z$ in the complex plane is a circle.

Let $ z = x + iy$

Now,

$\dfrac{{\bar z + 2}}{{\bar z - 1}} = \dfrac{{x - iy + 2}}{{x - iy - 1}}$

$= \dfrac{{[(x + 2) - iy][(x - 1) + iy]}}{{[(x - 1) - iy][(x - 1) + iy]}}$

$= \dfrac{{(x - 1)(x + 2) - iy(x - 1) + iy(x + 2) + {y^2}}}{{{{(x - 1)}^2} + {y^2}}}$

$= \dfrac{{(x - 1)(x + 2) + {y^2} + i[(x + 2)y - (x - 1)y]}}{{{{(x - 1)}^2} + {y^2}}} \left[ {\because  - {i^2} = 1} \right]$

${\text{Let,}}$

$\dfrac{{(x - 1)(x + 2) + {y^2}}}{{{{(x - 1)}^2} + {y^2}}} + \dfrac{{i[(x + 2)y - (x - 1)y]}}{{{{(x - 1)}^2} + {y^2}}} = x + iy$ 

Taking real part,

$\dfrac{{(x - 1)(x + 2) + {y^2}}}{{{{(x - 1)}^2} + {y^2}}} = 4$              (Given, that the real part = 4)

 $\Rightarrow {x^2} - x + 2x - 2 + {y^2} = 4\left( {{x^2} - 2x + 1 + {y^2}} \right)$

 $\Rightarrow {x^2} - x + 2x - 2 + {y^2} = 4{x^2} - 8x + 4 + 4{y^2}$ 

$\Rightarrow 3{x^2} + 3{y^2} - 9x + 6 = 0$, which represents a circle.

Hence, $z$ lies on the circle.

10. Show that the complex number $z$, satisfying the condition arg $\left( {\dfrac{{z - 1}}{{z + 1}}} \right) = \dfrac{\pi }{4}$ lies on a circle.

Ans: Given, The complex number $z$First use, $\arg \left( {\dfrac{{{z_1}}}{{{z_2}}}} \right) = \arg \left( {{z_1}} \right) - \arg \left( {{z_2}} \right)$. Also apply $\arg (z) = \theta  = {\tan ^{ - 1}}\dfrac{y}{x}$, where $z = x + iy$ and then use the property ${\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right)$

${\text{Let}},$

$z = x + iy$

${\text{Given that,}}$

 $\Rightarrow \arg (z - 1) - \arg (z + 1) = \dfrac{\pi }{4}$

 $\Rightarrow \arg (x + iy - 1) - \arg (x + iy + 1) = \dfrac{\pi }{4}$

 $\Rightarrow \arg (x - 1 + iy) - \arg (x + 1 + iy) = \dfrac{\pi }{4}$ 

$\Rightarrow  {\tan ^{ - 1}}\dfrac{y}{{x - 1}} - {\tan ^{ - 1}}\dfrac{y}{{x + 1}} = \dfrac{\pi }{4}$

$\Rightarrow  {\tan ^{ - 1}}\left[ {\dfrac{{\dfrac{y}{{x - 1}} - \dfrac{y}{{x + 1}}}}{{1 + \left( {\dfrac{y}{{x - 1}}} \right)\left( {\dfrac{y}{{x + 1}}} \right)}}} \right] = \dfrac{\pi }{4}$

$\Rightarrow  \dfrac{{y\left[ {\dfrac{{x + 1 - x + 1}}{{{x^2} - 1}}} \right]}}{{\dfrac{{{x^2} - 1 + {y^2}}}{{{x^2} - 1}}}} = \tan \dfrac{\pi }{4}$

$\Rightarrow  \dfrac{{2y}}{{{x^2} + {y^2} - 1}} = 1$

$\Rightarrow  {x^2} + {y^2} - 1 = 2y$

$\Rightarrow  {x^2} + {y^2} - 2y - 1 = 0$, which represents a circle.

11. Solve the equation $\left| z \right| = z + 1 + 2i$.

Ans: Given, $\left| z \right| = z + 1 + 2i$

Simplify $\left| z \right| = z + 1 + 2i$ by using the formula $|z| + \sqrt {{x^2} = {y^2}}$.

The given equation is $|z| = z + 1 + 2i.....\left( i \right)$

Let $ z = x + iy$

From Eq. (i), $ |x + iy| = x + iy + 1 + 2i$

 $\Rightarrow  \sqrt {{x^2} + {y^2}}  = x + iy + 1 + 2i \left[ {\because |z| + \sqrt {{x^2} = {y^2}} } \right]$

 $\Rightarrow  \sqrt {{x^2} + {y^2}}  = (x + 1) + i(y + 2)$ 

On squaring both sides, we get

${x^2} + {y^2} = {(x + 1)^2} + {i^2}{(y + 2)^2} + 2i(x + 1)(y + 2)$

 $\Rightarrow {x^2} + {y^2} = {x^2} + 2x + 1 - {y^2} - 4y - 4 + 2i(x + 1)(y + 2)$ 

On comparing real and imaginary parts,

${x^2} + {y^2} = {x^2} + 2x + 1 - {y^2} - 4y - 4$

${\text{i}}{\text{.e}}{\text{., }}2{y^2} = 2x - 4y - 3.....\left( {ii} \right)$ 

and

  $2(x + 1)(y + 2) = 0$

  $(x + 1) = 0{\text{ or }}(y + 2) = 0$

 $\Rightarrow x =  - 1{\text{ or }}y =  - 2$ 

For $x =  - 1$,

$2{y^2} =  - 2 - 4y - 3$[using Eq. (ii)]

$2{y^2} + 4y + 5 = 0$

 $\Rightarrow y = \dfrac{{ - 4 \pm \sqrt {16 - 2 \times 4 \times 5} }}{4}$

 $\Rightarrow y = \dfrac{{ - 4 \pm \sqrt { - 24} }}{4} \notin R$ 

${\text{For }}y =  - 2,$

  $2{( - 2)^2} = 2x - 4( - 2) - 3[{\text{using}}{\text{Eq}}.({\text{ii}})]$ 

 $\Rightarrow 8 = 2x + 8 - 3$

 $\Rightarrow 2x = 3$

 $\Rightarrow x = \dfrac{3}{2}$ 

$\therefore  z = x + iy = \dfrac{3}{2} - 2i$

12. If $|z + 1| = z + 2(1 + i)$, then find the value of $z$.

Ans: Given, $|z + 1| = z + 2(1 + i)$

Take $z = x + iy$ and use the formula $|z| + \sqrt {{x^2} = {y^2}}$.

Given that, $ |z + 1| = z + 2(1 + i).....\left( i \right)$

Let  $z = x + iy$

Then,

 $\Rightarrow |x + iy + 1| = x + iy + 2(1 + i)$

 $\Rightarrow |x + 1 + \mathop {iy}\limits^. | = (x + 2) + i(y + 2)$

 $\Rightarrow \sqrt {{{(x + 1)}^2} + {y^2}}  = (x + 2) + i(y + 2)$ 

On squaring both sides, we get

 $\Rightarrow {(x + 1)^2} + {y^2} = {(x + 2)^2} + {i^2}{(y + 2)^2} + 2i(x + 2)(y + 2)$

 $\Rightarrow {x^2} + 2x + 1 + {y^2} = {x^2} + 4x + 4 - {y^2} - 4y - 4 + 2i(x + 2)(y + 2)$

 $\Rightarrow {x^2} + {y^2} + 2x + 1 = {x^2} - {y^2} + 4x - 4y + 2i(x + 2)(y + 2)$ 

On comparing real and imaginary parts, we get

 $\Rightarrow {x^2} + {y^2} + 2x + 1 = {x^2} - {y^2} + 4x - 4y$

 $\Rightarrow 2{y^2} - 2x + 4y + 1 = 0.....\left( {ii} \right)$

${\text{and }}2(x + 2)(y + 2) = 0$

 $\Rightarrow x + 2 = 0{\text{ or }}y + 2 = 0$

 $\Rightarrow x =  - 2{\text{ or }}y =  - 2.....\left( {iii} \right)$ 

${\text{For  }}x =  - 2,$

  $2{y^2} + 4 + 4y + 1 = 0 \left[ {{\text{Using Eq}}{\text{. }}\left( {ii} \right)} \right]$

 $\Rightarrow 2{y^2} + 4y + 5 = 0$

 $\Rightarrow 16 - 4 \times 2 \times 5 < 0$ 

$\therefore$ Discriminant, $D = {b^2} - 4ac < 0$

$\Rightarrow  2{y^2} + 4y + 5$ has no real roots.

${\text{For  }}y =  - 2,$

 $\Rightarrow 2{( - 2)^2} - 2x + 4( - 2) + 1 = 0\left[ {{\text{Using Eq}}{\text{. }}\left( {ii} \right)} \right]$

 $\Rightarrow 8 - 2x - 8 + 1 = 0$

 $\Rightarrow x = \dfrac{1}{2}$

$\therefore  z = x + iy = \dfrac{1}{2} - 2i$ 

13. If arg $(z - 1) = arg(z + 3i)$, then find $x - 1:y$, where $z = x + iy$.

Ans: Given, arg $(z - 1) = \arg (z + 3i)$, where $z = x + iy$.

Apply $\arg (z) = \theta  = {\tan ^{ - 1}}\dfrac{y}{x}$, where $z = x + iy$.

Given that,

$\arg (z - 1) = \arg (z + 3i)$

Let  $z = x + iy$

${\text{Now,}}$

 $\Rightarrow \arg (x + \mathop {iy}\limits^.  - 1) = \arg (x + \mathop {iy}\limits^.  + 3i)$

 $\Rightarrow \arg (z - 1) = \arg (z + 3i)$

 $\Rightarrow \arg (x - 1 + \dot y) = \arg [x + i(y + 3)]$

 $\Rightarrow  {\tan ^{ - 1}}\dfrac{y}{{x - 1}} = {\tan ^{ - 1}}\dfrac{{y + 3}}{x}$

 $\Rightarrow  \dfrac{y}{{x - 1}} = \dfrac{{y + 3}}{x}$

 $\Rightarrow xy = (x - 1)(y + 3)$

 $\Rightarrow xy = xy - y + 3x - 3$

 $\Rightarrow 3x - 3 = y$

 $\Rightarrow \dfrac{{3\left( {x - 1} \right)}}{y} = 1$

 $\Rightarrow \dfrac{{x - 1}}{y} = \dfrac{1}{3}$

$\therefore \left( {x - 1} \right):y = 1:3$ 

14. Show that $\left| {\dfrac{{z - 2}}{{z - 3}}} \right| = 2$ represents a circle. Find its centre and radius.

Ans: Given, $\left| {\dfrac{{z - 2}}{{z - 3}}} \right| = 2$

If ${z_1} = {x_1} + i{y_1}$ and ${z_2} = {x_2} + i{y_2}$ are two complex numbers, then $\left| {\dfrac{{{z_1}}}{{{z_2}}}} \right| = \dfrac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}},\left( {{z_2} \ne 0} \right)$, use this concept to solve the above problem. Also, the general equation of a circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$, with centre $( - g, - f)$ and radius $= \sqrt {{g^2} + {f^2} - c}$.

Let $z = x + iy$

Given,

Equation is $\left| {\dfrac{{z - 2}}{{z - 3}}} \right| = 2$

$\Rightarrow \left| {\dfrac{{z - 2}}{{z - 3}}} \right| = 2$

$\Rightarrow \left| {\dfrac{{x + iy - 2}}{{x + iy - 3}}} \right| = 2$

$\Rightarrow |x - 2 + iy| = 2|x - 3 + iy|$

$\Rightarrow \sqrt {{{(x - 2)}^2} + {y^2}}  = 2\sqrt {{{(x - 3)}^2} + {y^2}}  \left[ {\because |x + iy| = \sqrt {{x^2} + {y^2}} } \right]$

On squaring both sides, we get

 $\Rightarrow {x^2} - 4x + 4 + {y^2} = 4\left( {{x^2} - 6x + 9 + {y^2}} \right)$

 $\Rightarrow 3{x^2} + 3{y^2} - 20x + 32 = 0$

 $\Rightarrow {x^2} + {y^2} - \dfrac{{20}}{3}x + \dfrac{{32}}{3} = 0.....\left( i \right)$ 

On comparing the above equation with ${x^2} + {y^2} + 2gx + 2fy + c = 0$, we get

 $\Rightarrow  2g = \dfrac{{ - 20}}{3}$

 $\Rightarrow g = \dfrac{{ - 10}}{3}$ 

and

 $\Rightarrow 2f = 0$

 $\Rightarrow f = 0{\text{ and }}c = \dfrac{{32}}{3}$

$\therefore {\text{ Centre }} = \left( { - g, - f} \right) = \left( {\dfrac{{10}}{3},0} \right)$ 

Also,

radius \[(r) = \sqrt {{{\left( {\dfrac{{10}}{3}} \right)}^2} + 0 - \dfrac{{32}}{3}}  \left[ {\because r = \sqrt {{g^2} + {f^2} - c} } \right]\]

$= \dfrac{1}{3}\sqrt {(100 - 96)}  = \dfrac{2}{3}$

15. If $\dfrac{{z - 1}}{{z + 1}}$ is a purely imaginary number , then find the value of $\left| z \right|$. 

Ans: Let,

$z = x + iy$

 $\Rightarrow \dfrac{{z - 1}}{{z + 1}} = \dfrac{{x + iy - 1}}{{x + iy + 1}}z \ne  - 1$

 $\Rightarrow \dfrac{{x - 1 + iy}}{{x + 1 + \mathop {iy}\limits^. }} = \dfrac{{(x - 1 + iy)(x + 1 - iy)}}{{(x + 1 + iy)(x + 1 - iy)}}$ 

${} { = \dfrac{{\left( {{x^2} - 1} \right) + iy(x + 1) - iy(x - 1) - {i^2}{y^2}}}{{{{(x + 1)}^2} - {{(iy)}^2}}}}$ 

${ \Rightarrow  }{\dfrac{{z - 1}}{{z + 1}} = \dfrac{{\left( {{x^2} - 1} \right) + {y^2} + i[y(x + 1) - y(x - 1)]}}{{{{(x + 1)}^2} + {y^2}}}}$ 

Given that, $\dfrac{{z - 1}}{{z + 1}}$ is a purely imaginary number.

Then,

 $\Rightarrow \dfrac{{\left( {{x^2} - 1} \right) + {y^2}}}{{{{(x + 1)}^2} + {y^2}}} = 0$

 $\Rightarrow  {x^2} - 1 + {y^2} = 0$

 $\Rightarrow {x^2} + {y^2} = 1$

 $\Rightarrow \sqrt {{x^2} + {y^2}}  = \sqrt 1 \left[ {\because |z| = \sqrt {{x^2} + {y^2}} } \right]$

$\therefore |z| = 1$ 

16. ${z_1}$ and ${z_2}$ are two complex numbers such that $\left| {{z_1}} \right| = \left| {{z_2}} \right|$ and $arg\left( {{z_1}} \right) + arg\left( {{z_2}} \right) = \pi$, then show that ${z_1} =  - {\bar z_2}$.

Ans: Given, $\left| {{z_1}} \right| = \left| {{z_2}} \right|$ and $\arg \left( {{z_1}} \right) + \arg \left( {{z_2}} \right) = \pi$

$\arg (z) = \theta  = {\tan ^{ - 1}}\dfrac{y}{x}$, where $z = x + iy$.

Let ${z_1} = {r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)$ and ${z_2} = {r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right)$ are two complex numbers.

Given that, $ \left| {{z_1}} \right| = \left| {{z_2}} \right|$

and $ \arg \left( {{z_1}} \right) + \arg \left( {{z_2}} \right) = \pi$

If $ \left| {{z_1}} \right| = \left| {{z_2}} \right|$

${r_1} = {r_2}$

and if $ \arg \left( {{z_1}} \right) + \arg \left( {{z_2}} \right) = \pi$

${\theta _1} + {\theta _2} = \pi$

${\theta _1} = \pi  - {\theta _2}$

Now, $ {z_1} = {r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)$

$\Rightarrow  {z_1} = {r_2}\left[ {\cos \left( {\pi  - {\theta _2}} \right) + i\sin \left( {\pi  - {\theta _2}} \right)} \right] \left[ {\because {r_1} = {r_2}} \right.$ and $\left. {{\theta _1} = \left( {\pi  - {\theta _2}} \right)} \right]$

$\Rightarrow  {z_1} = {r_2}\left( { - \cos {\theta _2} + i\sin {\theta _2}} \right)$

$\Rightarrow  {z_1} =  - {r_2}\left( {\cos {\theta _2} - i\sin {\theta _2}} \right)$

$\Rightarrow  {z_1} =  - \left[ {{r_2}\left( {\cos {\theta _2} - i\sin {\theta _2}} \right)} \right]$

$\Rightarrow  {z_1} =  - {\bar z_2}\left[ {\because {{\bar z}_2} = {r_2}\left( {\cos {\theta _2} - i\sin {\theta _2}} \right)} \right]$

17. If $\left| {{z_1}} \right| = 1\left( {{z_1} \ne  - 1} \right)$ and ${z_2} = \dfrac{{{z_1} - 1}}{{{z_1} + {1^\prime }}}$., then show that the real part of ${z_2}$ is zero.

Ans: Given, $\left| {{z_1}} \right| = 1\left( {{z_1} \ne  - 1} \right)$ and ${z_2} = \dfrac{{{z_1} - 1}}{{{z_1} + {1^\prime }}}$.

Take ${z_1} = x + iy$ and use the formula $\left| {z} \right| = \sqrt {{x^2} + {y^2}}$.

Let,

${z_1} = x + iy$

 $\Rightarrow \left| {{z_1}} \right| = \sqrt {{x^2} + {y^2}}  = 1 \left[ {\because \left| {{z_1}} \right| = 1,{\text{ given }}} \right].....\left( i \right)$ 

Now,

${z_2} = \dfrac{{{z_1} - 1}}{{{z_1} + 1}} = \dfrac{{x + iy - 1}}{{x + iy + 1}}$

$= \dfrac{{x - 1 + iy}}{{x + 1 + iy}} = \dfrac{{(x - 1 + iy)(x + 1 - iy)}}{{(x + 1 + iy)(x + 1 - iy)}}$

$= \dfrac{{{x^2} - 1 + iy(x + 1) - iy(x - 1) - {i^2}{y^2}}}{{{{(x + 1)}^2} - {i^2}{y^2}}}$

$= \dfrac{{{x^2} - 1 + ixy + \dot y - ixy + iy + {y^2}}}{{{{(x + 1)}^2} + {y^2}}}$

$= \dfrac{{{x^2} + {y^2} - 1 + 2iy}}{{{{\left( {x + 1} \right)}^2} + {y^2}}} = \dfrac{{1 - 1 + 2iy}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}\left[ {\because {x^2} + {y^2} = 1} \right]$

$= 0 + \dfrac{{2yi}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}$ 

Hence, the real part of \[{z_2}\] is zero.

18. If ${z_1},{z_2}$ and ${z_3},{z_4}$ are two pairs of conjugate complex numbers, then find $arg\left( {\dfrac{{{z_1}}}{{{z_4}}}} \right) + arg\left( {\dfrac{{{z_2}}}{{{z_3}}}} \right)$.

Ans: Given, ${z_1},{z_2}$ and ${z_3},{z_4}$ are two pairs of conjugate complex numbers.

First let, $z = r(\cos \theta  + i\sin \theta )$ then conjugate of $z$ i.e., $\bar z = r(\cos \theta  - i\sin \theta )$. Use the

property arg $\left( {\dfrac{{{z_1}}}{{{z_2}}}} \right) = \arg \left( {{z_1}} \right) - \arg \left( {{z_2}} \right)$.

Let ${z_1} = {r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)$

Then, 

  $ {z_2} = {{\bar z}_1}$

 $\Rightarrow {r_1}\left( {\cos {\theta _1} - i\sin {\theta _1}} \right) = {r_1}\left[ {\cos \left( { - {\theta _1}} \right) + \sin \left( { - {\theta _1}} \right)} \right]$ 

${\text{Also,}}$

${\text{Let }}{{\text{z}}_3} = {r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right),$

${\text{Then,}}$

${{\text{z}}_4} = {{\bar z}_3} = {r_2}\left( {\cos {\theta _2} - i\sin {\theta _2}} \right)$ 

$\arg \left( {\dfrac{{{z_1}}}{{{z_4}}}} \right) + \arg \left( {\dfrac{{{z_2}}}{{{z_3}}}} \right) = \arg \left( {{z_1}} \right) - \arg \left( {{z_4}} \right) + \arg \left( {{z_2}} \right) - \arg \left( {{z_3}} \right)$

${ = {\theta _1} - \left( { - {\theta _2}} \right) + \left( { - {\theta _1}} \right) - {\theta _2}}{[\because \arg (z) = \theta ]}$ 

${ = {\theta _1} + {\theta _2} - {\theta _1} - {\theta _2} = 0}$ 

19. If $\left| {{z_1}} \right| = \left| {{z_2}} \right| = L = \left| {{z_n}} \right| = 1$, then show that $\left| {{z_1} + {z_2} + {z_3} + L + {z_n}} \right| = \left| {\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_2}}} + \dfrac{1}{{{z_3}}} + L + \dfrac{1}{{{z_n}}}} \right|$.

Ans: Given, $\left| {{z_1}} \right| = \left| {{z_2}} \right| =  \cdots= \left| {{z_n}} \right| = 1$

Using formula ${\left| {z} \right|^2} = z.\bar z{\text{and }}\left| {{z_1} + {z_2} + {z_3}....{z_n}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right| + \left| {{z_3}} \right| + ....\left| {{z_n}} \right|$ 

${\text{ Given that,}}$

${\left| {{z_1}} \right| = \left| {{z_2}} \right| =  \cdots= \left| {{z_n}} \right| = 1}$ 

${ \Rightarrow } {{{\left| {{z_1}} \right|}^2} = {{\left| {{z_2}} \right|}^2} =  \cdots  = {{\left| {{z_n}} \right|}^2} = 1}$ 

 $\Rightarrow  {{z_1}{{\bar z}_1} = {z_2}{{\bar z}_2} = {z_3}{{\bar z}_3} =  \cdots  = {z_n}{{\bar z}_n} = 1}$ 

 $\Rightarrow { {z_1} = \dfrac{1}{{{{\bar z}_1}}},{z_2} = \dfrac{1}{{{{\bar z}_2}}} =  \cdots  = {z_n} = \dfrac{1}{{{{\bar z}_n}}}}$ 

Now, 

$\left| {{z_1} + {z_2} + {z_3} + {z_4} +  \cdots  + {z_n}} \right|$

 $= \left| {\dfrac{{{z_1}{{\bar z}_1}}}{{{{\bar z}_1}}} + \dfrac{{{z_2}{{\bar z}_2}}}{{{{\bar z}_2}}} + \dfrac{{{z_3}{{\bar z}_3}}}{{{{\bar z}_3}}} +  \cdots  + \dfrac{{{z_n}{{\bar z}_n}}}{{{{\bar z}_n}}}} \right| \left[ {\because {z_1}\bar z = 1,{\text{ where }}{z_1} = \dfrac{1}{z},{z_1} = \dfrac{{\bar z}}{{\bar z - \bar z}},{z_1} = \bar z} \right]$ 

$= \left| {\dfrac{{{{\left| {{z_1}} \right|}^2}}}{{{{\bar z}_1}}} + \dfrac{{{{\left| {{z_2}} \right|}^2}}}{{{{\bar z}_2}}} + \dfrac{{{{\left| {{z_3}} \right|}^2}}}{{{z_3}}} +  \cdots  + \dfrac{{{{\left| {{z_n}} \right|}^2}}}{{{{\bar z}_n}}}} \right|$

$= \left| {\dfrac{1}{{{{\bar z}_1}}} + \dfrac{1}{{{{\bar z}_2}}} + \dfrac{1}{{{{\bar z}_3}}} +  \cdots  + \dfrac{1}{{{{\bar z}_n}}}} \right| = \sqrt {\dfrac{1}{{{{\bar z}_1}}} + \dfrac{1}{{{{\bar z}_2}}} + \dfrac{1}{{{{\bar z}_3}}}}$

$= \left| {\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_2}}} +  \cdots  + \dfrac{1}{{{z_n}}}} \right|$

Hence proved.

20. If the complex numbers ${z_1}$ and ${z_2}$, arg $\left( {{z_1}} \right) - arg\left( {{z_2}} \right) = 0$, then show that $\left| {{z_1} - {z_2}} \right| = \left| {{z_1}} \right| - \left| {{z_2}} \right|$.

Ans: Given, arg $\left( {{z_1}} \right) - \arg \left( {{z_2}} \right) = 0$.

Apply $\arg (z) = \theta$, where $z = x + iy$ and $z = r(\cos \theta  + i\sin \theta )$.

Let,

${z_1} = {r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)$

${z_2} = {r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right)$

 $\Rightarrow \arg \left( {{z_1}} \right) = {\theta _1}{\text{ and }}\arg \left( {{z_2}} \right) = {\theta _2}$ 

Given that, 

$\arg \left( {{z_1}} \right) - \arg \left( {{z_2}} \right) = 0$

${\theta _1} - {\theta _2} = 0$

 $\Rightarrow {\theta _1} = {\theta _2}$ 

${z_2} = {r_2}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)\left[ {\because {\theta _1} = {\theta _2}} \right]$

${z_1} - {z_2} = \left( {{r_1}\cos {\theta _1} - {r_2}\cos {\theta _1}} \right) + i\left( {{r_1}\sin {\theta _1} - {r_2}\sin {\theta _1}} \right)$

$\left| {{z_1} - {z_2}} \right| = \sqrt {{{\left( {{r_1}\cos {\theta _1} - {r_2}\cos {\theta _1}} \right)}^2} + {{\left( {{r_1}\sin {\theta _1} - {r_2}\sin {\theta _1}} \right)}^2}}$

$= \sqrt {r_1^2 + r_2^2 - 2{r_1}{r_2}{{\cos }^2}{\theta _1} - 2{r_1}{r_2}{{\sin }^2}{\theta _1}}$

$= \sqrt {r_1^2 + r_2^2 - 2{q_1}{r_2}\left( {{{\sin }^2}{\theta _1} + {{\cos }^2}{\theta _1}} \right)}$

$= \sqrt {r_1^2 + r_2^2 - 2{r_1}{r_2}}  = \sqrt {{{\left( {{r_1} - {r_2}} \right)}^2}}$

$\left| {{z_1} - {z_2}} \right| = {r_1} - {r_2}  [\because r = |z|]$

 $= \left| {{z_1}} \right| - \left| {{z_2}} \right|$ 

Hence proved.

21. Solve the system of equations $Re\left( {{z^2}} \right) = 0,|z| = 2$.

Ans: Given, $\operatorname{Re} \left( {{z^2}} \right) = 0,|z| = 2$

Let $z = x + iy$ and use the formula $|z| = \sqrt {{x^2} + {y^2}}$.

Given that, $ \operatorname{Re} \left( {{z^2}} \right) = 0,|z| = 2$

Let $ z = x + iy$

$|z| = \sqrt {{x^2} + {y^2}}$

$\therefore$$\sqrt {{x^2} + {y^2}}  = 2$                                        (On squaring both sides)

$\Rightarrow{x^2} + {y^2} = 4.....\left( i \right)$

and $\operatorname{Re} (z) = x$

Also, $ z = x + iy$

$\Rightarrow{z^2} = {x^2} + 2ixy - {y^2}$                          (On squaring both sides)

$\Rightarrow{z^2} = \left( {{x^2} - {y^2}} \right) + 2ixy$

$\Rightarrow\operatorname{Re} \left( {{z^2}} \right) = {x^2} - {y^2}\left[ {\because \operatorname{Re} \left( {{z^2}} \right) = 0} \right]$

$\Rightarrow{x^2} - {y^2} = 0.....\left( {ii} \right)$

On adding Eq. (i) and (ii), 

${x^2} + {y^2} + {x^2} - {y^2} = 4 + 0$

 $\Rightarrow 2{x^2} = 4$

 $\Rightarrow {x^2} = 2$

 $\Rightarrow x =  \pm \sqrt 2$ 

Putting the value of $x$ in equation (i), we get

$y =  \pm \sqrt 2$

$\because$ $z = x + iy$

$\Rightarrow z = \sqrt 2  \pm i\sqrt 2 , - \sqrt 2  \pm i\sqrt 2$

22. Find the complex number satisfying the equation $z + \sqrt 2 |(z + 1)| + i = 0$.

Ans: Given, $z + \sqrt 2 |(z + 1)| + i = 0$

Let $z = x + iy$ and use the formula $|z| = \sqrt {{x^2} + {y^2}}$. Compare the real and imaginary parts of LHS and RHS.

Given equation is $ z + \sqrt 2 |(z + 1)| + i = 0$                       …..(i)

Let $ z = x + iy$

On substituting the value of $z$ in equation (i), we get

$\Rightarrow  x + iy + \sqrt 2 |x + \dot y + 1| + i = 0$

$\Rightarrow  x + i(1 + y) + \sqrt 2 \left[ {\sqrt {{{(x + 1)}^2} + {y^2}} } \right] = 0$

$\Rightarrow  x + i(1 + y) + \sqrt 2 \sqrt {\left( {{x^2} + 2x + 1 + {y^2}} \right)}  = 0$

On comparing real and imaginary parts of LHS with RHS, we get

$\Rightarrow  x + \sqrt 2 \sqrt {{x^2} + 2x + 1 + {y^2}}  = 0$       (real part)

 $\Rightarrow {x^2} = 2\left( {{x^2} + 2x + 1 + {y^2}} \right)$

 $\Rightarrow {x^2} + 4x + 2{y^2} + 2 = 0.....\left( {ii} \right)$ 

$1 + y = 0$                                                 (imaginary part)

$\Rightarrow y =  - 1$

For $y =  - 1,$

$\Rightarrow {x^2} + 4x + 2 + 2 = 0\left[ {{\text{Using eq}}{\text{. }}\left( {ii} \right)} \right]$

 $\Rightarrow {x^2} + 4x + 4 = 0$

 $\Rightarrow {(x + 2)^2} = 0$

 $\Rightarrow x + 2 = 0$

 $\Rightarrow x =  - 2$

$\therefore z = x + iy =  - 2 - i$ 

23. Write the complex number $z = \dfrac{{1 - i}}{{cos\dfrac{\pi }{3} + isin\dfrac{\pi }{3}}}$ in polar form.

Ans: Given, $z = \dfrac{{1 - i}}{{\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}}}$

Use trigonometrically identity to convert complex number into polar form.

Given that,

$z = \dfrac{{1 - i}}{{\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}}}\left( {{\text{Multiplying and dividing numerator by }}\sqrt 2 } \right)$

 $\Rightarrow \dfrac{{ - \sqrt 2 \left[ {\dfrac{{ - 1}}{{\sqrt 2 }} + i\dfrac{1}{{\sqrt 2 }}} \right]}}{{\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}}} = \dfrac{{ - \sqrt 2 \left[ {\cos \left( {\pi  - \dfrac{\pi }{4}} \right) + i\sin \left( {\pi  - \dfrac{\pi }{4}} \right)} \right]}}{{\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}}}$

 $\Rightarrow \dfrac{{ - \sqrt 2 \left[ {\cos \dfrac{{3\pi }}{4} + i\sin \dfrac{{3\pi }}{4}} \right]}}{{\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}}}$ 

$\Rightarrow  - \sqrt 2 \left[ {\cos \left( {\dfrac{{3\pi }}{4} - \dfrac{\pi }{3}} \right) + i\sin \left( {\dfrac{{3\pi }}{4} - \dfrac{\pi }{3}} \right)} \right]$

$\Rightarrow  - \sqrt 2 \left[ {\cos \dfrac{{5\pi }}{{12}} + i\sin \dfrac{{5\pi }}{{12}}} \right]$

24. If $z$ and $w$ are two complex numbers such that $|zw| = 1$ and $arg(z) - arg(w) = \dfrac{\pi }{2}$, then show that $\bar zw =  - i$.

Ans: Given, $z$ and $w$ are two complex numbers such that $|zw| = 1$ and $\arg (z) - \arg (w) = \dfrac{\pi }{2}$.

Let $z = {r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)$ and $w = {r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right)$and  $|zw| = |z||w|$.

Let $ z = {r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)$ and $w = {r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right)$

Also, $ |zw| = |z||w| = {r_1}{r_2} = 1$

$\therefore {r_1}{r_2} = 1$

Further, $ \arg (z) = {\theta _1}$ and $\arg (w) = {\theta _2}$

${\text{But,}}$

  $\arg (z) - \arg (w) = \dfrac{\pi }{2}$

 $\Rightarrow {\theta _1} - {\theta _2} = \dfrac{\pi }{2}$

 $\Rightarrow \arg \left( {\dfrac{z}{w}} \right) = \dfrac{\pi }{2}$ 

Now, 

to prove $\bar zw =  - i$

${\text{LHS}} = \bar zw$

$= {r_1}\left( {\cos {\theta _1} - i\sin {\theta _1}} \right){r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right)$

$= {r_1}{r_2}\left[ {\cos \left( {{\theta _2} - {\theta _1}} \right) + i\sin \left( {{\theta _2} - {\theta _1}} \right)} \right]$

 $= \left[ {{r_1}{r_2}\cos \left( {\dfrac{{ - \pi }}{2}} \right) + i\sin \left( {\dfrac{{ - \pi }}{2}} \right)} \right]$

$= 1[0 - i] =  - i = {\text{RHS}}$ 

25. Fill in the blanks.

(i) For any two complex numbers ${z_1},{z_2}$ and any real numbers $a, b$, ${\left| {a{z_1} - b{z_2}} \right|^2} + {\left| {b{z_1} + a{z_2}} \right|^2} = .....$

Ans: ${\left| {a{z_1} - b{z_2}} \right|^2} + {\left| {b{z_1} + a{z_2}} \right|^2}$

$= {\left| {a{z_1}} \right|^2} + {\left| {b{z_2}} \right|^2} - 2\operatorname{Re} \left( {a{z_1}.b{{\bar z}_2}} \right) + {\left| {b{z_1}} \right|^2} + {\left| {a{z_2}} \right|^2} + 2\operatorname{Re} \left( {a{z_1}.b{{\bar z}_2}} \right)$

 $= \left( {{a^2} + {b^2}} \right){\left| {{z_1}} \right|^2} + \left( {{a^2} + {b^2}} \right){\left| {{z_2}} \right|^2}$

 $= \left( {{a^2} + {b^2}} \right)\left( {{{\left| {{z_1}} \right|}^2} + {{\left| {{z_2}} \right|}^2}} \right)$ 

(ii) The value of $\sqrt { - 25}  \times \sqrt { - 9}$ is ...

Ans:  ${\text{ }}\sqrt { - 25}  \times \sqrt { - 9}$

 $\Rightarrow i\sqrt {25}  \times i\sqrt 9  = {i^2}(5 \times 3)$

 $\Rightarrow  - 15$ 

(iii) The number $\dfrac{{{{(1 - i)}^3}}}{{1 - {i^3}}}$ is equal to ...

Ans: ${\text{ }}\dfrac{{{{(1 - i)}^3}}}{{1 - {i^3}}} = \dfrac{{{{(1 - i)}^3}}}{{(1 - i)\left( {1 + i + {i^2}} \right)}}$

 $\Rightarrow \dfrac{{{{(1 - i)}^2}}}{i} = \dfrac{{1 + {i^2} - 2i}}{i}$

 $\Rightarrow \dfrac{{ - 2i}}{i} =  - 2$ 

(iv) The sum of the series $i + {i^2} + {i^3} + ....$ upto 1000 terms is $\ldots$

Ans: $i + {i^2} + {i^3} +  \ldots$ upto $1000$ terms,

$= i + {i^2} + {i^3} + {i^4} +  \ldots {i^{1000}} = 0$

$\left[ {\because {i^n} + {i^{n + 1}} + {i^{n + 2}} + {i^{n + 3}} = 0} \right.$, where $n \in N$ i.e., $\left. {\sum\limits_{n = 1}^{1000} {{i^n}}  = 0} \right]$

(v) Multiplicative inverse of $1 + i$ is $ \ldots $

Ans: Multiplicative inverse of $1 + i$

$\dfrac{1}{{1 + i}} = \dfrac{{1 - i}}{{1 - {i^2}}}$

 $\Rightarrow \dfrac{1}{2}(1 - i)$ 

(vi) If ${z_1}$ and ${z_2}$ are complex numbers such that ${z_1} + {z_2}$ is a real number, then ${z_1} = ...$

Ans: Let ${z_1} = {x_1} + i{y_1}$ and ${z_2} = {x_2} + i{y_2}$

${z_1} + {z_2} = \left( {{x_1} + {x_2}} \right) + i\left( {{y_1} + {y_2}} \right)$ which is real.

If ${z_1} + {z_2}$ is real, then ${y_1} + {y_2} = 0$

$\Rightarrow  {y_1} =  - {y_2}$

$\because  {z_2} = {x_2} - {\dot i_1}$

$\Rightarrow {z_2} = {\bar z_1}\left[ {{\text{when }}{x_1} = {x_2}} \right]$

(vii) $arg(z) + arg\bar z$ where,  is $ \ldots $

Ans: $\arg (z) + \arg (\bar z),(\bar z \ne 0)$

$\Rightarrow  \theta  + ( - \theta ) = 0$

(viii) If $|z + 4|\leqslant3$, then the greatest and least values of $|z + 1|$ are ... and ….

Ans: Given that, $|z + 4| \leqslant 3$ For the greatest value of $|z + 1|$.

 $\Rightarrow |z + 1| = |z + 4 - 3| \leqslant |z + 4| + | - 3|$

 $\Rightarrow |z + 4 - 3| \leqslant 3 + 3$

 $\Rightarrow |z + 4 - 3| \leqslant 6$ 

So, greatest value of $|z + 1| = 6$ For, now, least value of $|z + 1|$, we know that the least value of the modulus of a complex number is zero. So, the least value of $|z + 1|$ is zero.

(ix) If $\left| {\dfrac{{z - 2}}{{z + 2}}} \right| = \dfrac{\pi }{6}$, then the locus of $z$ is ...

Ans: Given that, $\left| {\dfrac{{z - 2}}{{z + 2}}} \right| = \dfrac{\pi }{6}$

 $\Rightarrow  \dfrac{{|x + \dot y - 2|}}{{|x + iy + 2|}} = \dfrac{\pi }{6}$

 $\Rightarrow \dfrac{{|x - 2 + \dot iy|}}{{|x + 2 + \dot i|\mid }} = \dfrac{\pi }{6}$ 

$\Rightarrow  6|x - 2 + \mathop {iy}\limits^. | = \pi |x + 2 + iy|$

$\Rightarrow  6\sqrt {{{(x - 2)}^2} + {y^2}}  = \pi \sqrt {{{(x + 2)}^2} + {y^2}}$

$\Rightarrow  36\left[ {{x^2} + 4 - 4x + {y^2}} \right] = {\pi ^2}\left[ {{x^2} + 4x + 4 + {y^2}} \right]$

$\Rightarrow \left( {36 - {\pi ^2}} \right){x^2} + \left( {36 - {\pi ^2}} \right){y^2} - \left( {144 + 4{\pi ^2}} \right)x + 144 + 4{\pi ^2} = 0$, which is a circle.

(x) If $|z| = 4$ and $arg(z) = \dfrac{{5\pi }}{6}$, then $z = ...$

Ans: Given that, $|z| = 4$ and arg $(z) = \dfrac{{5\pi }}{6}$

${\text{Let,}}$

$z = x + iy = r(\cos \theta  + i\sin \theta )$

$\left| z \right| = r = 4{\text{ and }}\arg (z) = \theta$

 $\because \tan \theta  = \dfrac{{5\pi }}{6}$

 $\Rightarrow z = 4\left( {\cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}} \right)$

 $\Rightarrow 4\left[ {\cos \left( {\pi  - \dfrac{\pi }{6}} \right) + i\sin \left( {\pi  - \dfrac{\pi }{6}} \right)} \right]$

 $\Rightarrow 4\left[ { - \cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}} \right]$

 $\Rightarrow 4\left[ {\dfrac{{ - \sqrt 3 }}{2} + i\dfrac{1}{2}} \right] =  - 2\sqrt 3  + 2i$ 

26. State true or false for the following.

(i) The order relation is defined on the set of complex numbers.

Ans: False

We can compare two complex numbers when they are purely real. Otherwise comparison of complex number is not possible.

(ii) Multiplication of a non-zero complex number by $- i$ rotates the point about origin through a right angle in the anti-clockwise direction.

Ans: False

(iii) For any complex number $z$, the minimum value of $|z| + |z - 1|$ is 1.

Ans: True

 Let, $z = x + iy$

 $\Rightarrow |z| + |z - 1| = \sqrt {{x^2} + {y^2}}  + \sqrt {{{(x - 1)}^2} + {y^2}}$ 

If $x = 0,y = 0$, then the value of $|z| + |z - 1| = 1$.

(iv) The locus represented by $|z - 1| = |z - i|$ is a line perpendicular to the join of the points $(1,0)$ and $(0,1)$.

Ans: True

Let, $z = x + iy$

  $|z - 1| = |z - i|$ 

Then,

 $\Rightarrow |x - 1 + iy| = |x - i(1 - y)|$

 $\Rightarrow {(x - 1)^2} + {y^2} = {x^2} + {(1 - y)^2}$

 $\Rightarrow {x^2} - 2x + 1 + {y^2} = {x^2} + 1 + {y^2} - 2y$

 $\Rightarrow  - 2x + 1 = 1 - 2y$

 $\Rightarrow  - 2x + 2y = 0$

 $\Rightarrow x - y = 0$ 

Equation of a line through the points $(1,0)$ and $(0,1)$,                             …..(i)

$y - 0 = \dfrac{{1 - 0}}{{0 - 1}}(x - 1)$

$\Rightarrow y =  - (x - 1) \Rightarrow x + y = 1$                                                                           …..(ii)

which is perpendicular to the line $x - y = 0$.

(v) If $z$ is a complex number such that  and $Re(z) = 0$, then, $Im\left( {{z^2}} \right) = 0$

Ans: False

Let $z = x + iy,z \ne 0$ and $\operatorname{Re} (z) = 0$

i.e., $ x = 0$

$\therefore  z = iy$

$\operatorname{Im} \left( {{z^2}} \right) = {i^2}{y^2} =  - {y^2}$ which is real.

(vi) The inequality $|z - 4| < |z - 2|$ represents the region given by $x > 3$.

Ans: True

 Given inequality, $ |z - 4| < |z - 2|$

Let $ z = x + iy$

$\therefore {|x - 4 + \dot y| < |x - 2 + iy|}$ 

 $\Rightarrow {\sqrt {{{(x - 4)}^2} + {y^2}}  < \sqrt {{{(x - 2)}^2} + {y^2}} } $

$\Rightarrow  {(x - 4)^2} + {y^2} < {(x - 2)^2} + {y^2}$

$\Rightarrow  {x^2} - 8x + 16 + {y^2} < {x^2} - 4x + 4 + {y^2}$

 $\Rightarrow  - 8x + 16 <  - 4x + 4$

 $\Rightarrow  - 8x <  - 4x - 12$

 $\Rightarrow  - 4x <  - 12$

 $\Rightarrow 4x > 12$

 $\Rightarrow  x > 3$ 

(vii) Let ${z_1}$ and ${z_2}$ be two complex numbers such that $\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$, then arg $\left( {{z_1} - {z_2}} \right) = 0$

Ans: False

Let, ${z_1} = {x_1} + i{y_1} {\text{ and }} {z_2} = {x_2} + i{y_2}$

${\text{Given that,}}$

$\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$

 $\Rightarrow \left| {{x_1} + i{y_1} + {x_2} + i{y_2}} \right| = \left| {{x_1} + i{y_1}} \right| + \left| {{x_2} + i{y_2}} \right|$

 $\Rightarrow \sqrt {{{\left( {{x_1} + {x_2}} \right)}^2} + {{\left( {{y_1} + {y_2}} \right)}^2}}  = \sqrt {\left( {{x_1}^2 + {y_1}^2} \right)}  + \sqrt {\left( {{x_2}^2 + {y_2}^2} \right)}$ 

On squaring both sides, we get

${\left( {{x_1} + {x_2}} \right)^2} + {\left( {{y_1} + {y_2}} \right)^2} = \left( {x_1^2 + y_1^2} \right) + \left( {x_2^2 + y_2^2} \right) + 2\sqrt {\left( {x_1^2 + y_1^2} \right)\left( {x_2^2 + y_2^2} \right)}$

$\Rightarrow x_1^2 + x_2^2 + 2{x_1}{x_2} + y_1^2 + y_2^2 + 2{y_1}{y_2} = x_1^2 + y_1^2 + x_2^2 + y_2^2 + 2\sqrt {\left( {x_1^2 + y_1^2} \right)\left( {x_2^2 + y_2^2} \right)}$

 $\Rightarrow {} {2{x_1}{x_2} + 2{y_1}{y_2}}{ = 2\sqrt {\left( {x_1^2 + y_1^2} \right)\left( {x_2^2 + y_2^2} \right)} }$ 

 $\Rightarrow {}{{x_1}{x_2} + {y_1}{y_2}}{ = \sqrt {\left( {x_1^2 + y_1^2} \right)\left( {x_2^2 + y_2^2} \right)} }$ 

On squaring both sides, we get 

  $x_1^2x_2^2 + y_1^2y_2^2 + 2{x_1}{x_2}{y_1}{y_2} = x_1^2x_2^2 + y_1^2x_2^2 + x_1^2y_2^2 + y_1^2y_2^2$

 $\Rightarrow {\left( {{x_1}{y_2} - {x_2}{y_1}} \right)^2} = 0$ 

$\Rightarrow  {x_1}{y_2} = {x_2}{y_1}$

$\Rightarrow  \dfrac{{{y_1}}}{{{x_1}}} = \dfrac{{{y_2}}}{{{x_2}}}$

$\Rightarrow  \left( {\dfrac{{{y_1}}}{{{x_1}}}} \right) - \left( {\dfrac{{{y_2}}}{{{x_2}}}} \right) = 0$

$\Rightarrow  \arg \left( {{z_1}} \right) - \arg \left( {{z_2}} \right) = 0$

(viii) 2 is not a complex number.

Ans: True

We know that, 2 is a real number.

Since, 2 is not a complex number.

27. Match the statements of Column A and Column B.

Ans: (i) Given,

$z = i + \sqrt 3  = r(\cos \theta  + \sin \theta )$

$\because r\cos \theta  = \sqrt 3 ,r\sin \theta  = 1$

 $\Rightarrow {r^2} = 1 + 3 = 4\left[ {\because r > 0} \right]$

 $\Rightarrow r = 2$ 

$\Rightarrow  \tan \alpha= \left| {\dfrac{{r\sin \theta }}{{r\cos \theta }}} \right| = \dfrac{1}{{\sqrt 3 }}$

 $\Rightarrow  \tan \alpha  = \dfrac{1}{{\sqrt 3 }}$

 $\Rightarrow \alpha  = \dfrac{\pi }{6}$ 

$\because x > 0, y > 0$

and $ \arg (z) = \theta  = \dfrac{\pi }{6}$

So the polar form of $z$ is $2\left( {\cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}} \right)$.

(ii) Given that,

$z =  - 1 + \sqrt { - 3}  =  - 1 + i\sqrt 3$

$\therefore  \tan \alpha= \left| {\dfrac{{\sqrt 3 }}{{ - 1}}} \right| = \sqrt 3$

\[ \Rightarrow \tan \alpha  = \tan \dfrac{\pi }{3}\]

 $\Rightarrow \alpha  = \dfrac{\pi }{3}$

  $\because x < 0,y > 0$

 $\Rightarrow \theta  = \pi  - \alpha$

$= \pi  - \dfrac{\pi }{3} = \dfrac{{2\pi }}{3}$ 

(iii) Given that,

$|z + 2| = |z - 2|$

$\Rightarrow  |x + 2 + \dot r| = |x - 2 + iy|$

$\Rightarrow  {(x + 2)^2} + {y^2} = {(x - 2)^2} + {y^2}$

 $\Rightarrow  {x^2} + 4x + 4 = {x^2} - 4x + 4$

 $\Rightarrow 8x = 0$

$\therefore x = 0$ 

It is a straight line which is a perpendicular bisector of segment joining the points $( - 2,0)$ and $(2,0)$.

(iv) Given that,

$|z + 2i| = |z - 2i|$

$\Rightarrow  |x + i(y + 2)| = |x + i(y - 2)|$

$\Rightarrow {x^2} + {(y + 2)^2} = {x^2} + {(y - 2)^2}$

 $\Rightarrow  4y = 0$

 $\Rightarrow y = 0$ 

It is a straight line, which is a perpendicular bisector of segment joining $(0, - 2)$ and $(0,2)$.

(v) Given that,

  $|z + 4i| \geqslant 3 = |x + iy + 4i| \geqslant 3$

 $\Rightarrow |x + i(y + 4)| \geqslant 3$

 $\Rightarrow \sqrt {{x^2} + {{(y + 4)}^2}}  \geqslant 3$

 $\Rightarrow {x^2} + {(y + 4)^2} \geqslant 9$

 $\Rightarrow {x^2} + {y^2} + 8y + 16 \geqslant 9$

 $\Rightarrow {x^2} + {y^2} + 8y + 7 \geqslant 0$ 

Which represents a circle. On or outside having center $(0, - 4)$ and radius 3.

(vi) Given that, 

$|z + 4| \leqslant 3$

 $\Rightarrow |x + iy + 4| \leqslant 3$

 $\Rightarrow |x + 4 + iy| \leqslant 3$

 $\Rightarrow \sqrt {{{(x + 4)}^2} + {y^2}}  \leqslant 3$ 

$\Rightarrow {(x + 4)^2} + {y^2} \leqslant 9$

$\Rightarrow  {x^2} + 8x + 16 + {y^2} \leqslant 9$

$\Rightarrow  {x^2} + 8x + {y^2} + 7 \leqslant 0$

It represents the region which is on or inside the circle having the center $( - 4,0)$ and radius 3.

(vii) Given that,

$z = \dfrac{{1 + 2i}}{{1 - i}} = \dfrac{{(1 + 2i)(1 + i)}}{{(1 - i)(1 + i)}}$

 $\Rightarrow \dfrac{{1 + 2i + i + 2{i^2}}}{{1 - {i^2}}} = \dfrac{{1 - 2 + 3i}}{{1 + 1}}$

 $\Rightarrow \dfrac{{ - 1 + 3i}}{2}$

$\therefore \bar z = \dfrac{{ - 1}}{2} - \dfrac{{3i}}{2}$ 

Hence, $\left( {\dfrac{{ - 1}}{2},\dfrac{{ - 3}}{2}} \right)$ lies in the third quadrant.

(viii) Given that,

  $z = 1 - i$

$\therefore \dfrac{1}{z} = \dfrac{1}{{1 - i}}$ 

 $\Rightarrow \dfrac{{1 + i}}{{\left( {1 - i} \right)\left( {1 + i} \right)}} = \dfrac{{1 + i}}{{1 - {i^2}}}$

 $\Rightarrow \dfrac{1}{2}\left( {1 + i} \right)$ 

Hence, $\left( {\dfrac{1}{2},\dfrac{1}{2}} \right)$ lies in first quadrant.

28. What is the conjugate of $\dfrac{{2 - i}}{{{{\left( {1 - 2i} \right)}^2}}}$?

Ans: Given, $\dfrac{{2 - i}}{{{{\left( {1 - 2i} \right)}^2}}}$

If $z = x + iy{\text{ and }}\bar z = x - iy$.

Given that,

  $z = \dfrac{{2 - i}}{{{{(1 - 2i)}^2}}}$

 $\Rightarrow \dfrac{{2 - i}}{{1 + 4{i^2} - 4i}} = \dfrac{{2 - i}}{{1 - 4 - 4i}}$

 $\Rightarrow \dfrac{{2 - i}}{{ - 3 - 4i}} = \dfrac{{(2 - i)}}{{ - (3 + 4i)}}$

 $\Rightarrow  - \left[ {\dfrac{{(2 - i)(3 - 4i)}}{{(3 + 4i)(3 - 4i)}}} \right] =  - \left( {\dfrac{{6 - 8i - 3i + 4{i^2}}}{{9 + 16}}} \right)$

 $\Rightarrow  - \dfrac{{( - 11i + 2)}}{{25}} = \dfrac{{ - 1}}{{25}}(2 - 11i)$

 $\Rightarrow z = \dfrac{1}{{25}}( - 2 + 11i)$

 $\Rightarrow \bar z = \dfrac{1}{{25}}( - 2 - 11i) = \dfrac{{ - 2}}{{25}} - \dfrac{{11}}{{25}}i$ 

29. If $\left| {{z_1}} \right| = \left| {{z_2}} \right|$, is it necessary that ${z_1} = {z_2}$.

Ans: Given, $\left| {{z_1}} \right| = \left| {{z_2}} \right|$

If $z = x + iy$ then $\left| z \right| = \sqrt {{x^2} + {y^2}}$.

Given that,

$\left| {{z_1}} \right| = \left| {{z_2}} \right|$

${\text{Let,}}$

${z_1} = {x_1} + i{y_1}{\text{ and }}{z_2} = {x_2} + i{y_2}$

 $\Rightarrow \left| {{x_1} + i{y_1}} \right| = \left| {{x_2} + i{y_2}} \right|$

 $\Rightarrow x_1^2 + y_1^2 = x_2^2 + y_2^2$

 $\Rightarrow x_1^2 = x_2^2,y_1^2 = y_2^2$

 $\Rightarrow {x_1} =  \pm {x_2},{y_1} =  \pm {y_2}$

 $\Rightarrow {z_1} = {x_1} + i{y_1}{\text{ or }}{z_1} =  \pm {x_2} \pm i{y_2}$ 

Hence, it is not necessary that ${z_1} = {z_2}$.

30. If $\dfrac{{{{\left( {{a^2} + 1} \right)}^2}}}{{2a - i}} = x + iy$, then what is the value of ${x^2} + {y^2}$?

Ans: Given, $\dfrac{{{{\left( {{a^2} + 1} \right)}^2}}}{{2a - i}} = x + iy$

Rationalize the given expression and compare the real and imaginary parts.

Given that,

$\dfrac{{{{\left( {{a^2} + 1} \right)}^2}}}{{2a - i}} = x + iy$

 $\Rightarrow \dfrac{{{{\left( {{a^2} + 1} \right)}^2}}}{{2a - i}} \times \dfrac{{2a + i}}{{2a + i}} = x + iy$

 $\Rightarrow \dfrac{{{{\left( {{a^2} + 1} \right)}^2}\left( {2a + i} \right)}}{{4{a^2} + 1}} = x + iy$

 $\Rightarrow \dfrac{{2a{{\left( {{a^2} + 1} \right)}^2} + i{{\left( {a + 1} \right)}^2}}}{{4{a^2} + 1}} = x + iy$ 

On comparing the real and imaginary parts, we get

 $\Rightarrow x = \dfrac{{2a{{\left( {{a^2} + 1} \right)}^2}}}{{4{a^2} + 1}}{\text{  and  }}y = \dfrac{{{{\left( {{a^2} + 1} \right)}^2}}}{{4{a^2} + 1}}$

$\therefore {x^2} + {y^2} = 4{a^2}{\left[ {\dfrac{{{{\left( {{a^2} + 1} \right)}^2}}}{{4{a^2} + 1}}} \right]^2} + {\left[ {\dfrac{{{{\left( {{a^2} + 1} \right)}^2}}}{{4{a^2} + 1}}} \right]^2}$

$= \dfrac{{\left( {4{a^2} + 1} \right){{\left( {{a^2} + 1} \right)}^4}}}{{{{\left( {4{a^2} + 1} \right)}^2}}} = \dfrac{{{{\left( {{a^2} + 1} \right)}^4}}}{{\left( {4{a^2} + 1} \right)}}$ 

31. Find the value of $z$, if $\left| z \right| = 4$ and $arg\left( z \right) = \dfrac{{5\pi }}{6}$.

Ans: Given, $\left| z \right| = 4$ and $\arg \left( z \right) = \dfrac{{5\pi }}{6}$

If $z = x + iy$ then it’s polar form is $z = r(\cos \theta  + i\sin \theta )$ where $\arg \left( z \right) = \theta$.

Let, 

  $z = r(\cos \theta  + i\sin \theta )$

${\text{also   | }}z| = r = 4{\text{ and }}\theta  = \dfrac{{5\pi }}{6}\left[ {\because \arg \left( z \right) = \theta } \right]$

$\therefore z = 4\left[ {\cos \dfrac{{5\pi }}{6} + i\sin \dfrac{{5\pi }}{6}} \right]\left[ {\because z = r\left( {\cos \theta  + i\sin \theta } \right)} \right]$ 

 $\Rightarrow 4\left[ {\cos \left( {\pi  - \dfrac{\pi }{6}} \right) + i\sin \left( {\pi  - \dfrac{\pi }{6}} \right)} \right]$

 $\Rightarrow 4\left[ { - \cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}} \right]$

 $\Rightarrow 4\left[ { - \dfrac{{\sqrt 3 }}{2} + i\dfrac{1}{2}} \right] =  - 2\sqrt 3  + 2i$ 

32. Find the value of $\left| {(1 + i)\dfrac{{(2 + i)}}{{(3 + i)}}} \right|$

Ans: Given, $\left| {(1 + i)\dfrac{{(2 + i)}}{{(3 + i)}}} \right|$

First, convert the given expression in the formula $a + ib$, then use $\left| {a + ib} \right| = \sqrt {{a^2} + {b^2}}$.

Given that, 

$\left| {(1 + i)\dfrac{{(2 + i)}}{{(3 + i)}}} \right|$

 $\Rightarrow \left| {\dfrac{{\left( {2 + i + 2i + {i^2}} \right)}}{{(3 + i)}}} \right| = \left| {\dfrac{{2 + 3i - 1}}{{3 + i}}} \right|$

 $\Rightarrow \left| {\dfrac{{1 + 3i}}{{3 + i}}} \right| = \left| {\dfrac{{(1 + 3i)(3 - i)}}{{(3 + i)(3 - i)}}} \right|$ 

 $\Rightarrow \left| {\dfrac{{3 + 9i - i - 3{i^2}}}{{9 - {i^2}}}} \right| = \left| {\dfrac{{3 + 8i + 3}}{{9 + 1}}} \right|$

 $\Rightarrow \left| {\dfrac{{6 + 8i}}{{10}}} \right| = \sqrt {\dfrac{{{6^2}}}{{100}} + \dfrac{{{8^2}}}{{100}}}$

 $\Rightarrow \sqrt {\dfrac{{36 + 64}}{{100}}}  = \sqrt {\dfrac{{100}}{{100}}}  = 1$ 

33. Find the principal argument of ${\left( {1 + i\sqrt 3 } \right)^2}$

Ans: Given, ${\left( {1 + i\sqrt 3 } \right)^2}$

Let $z = a + ib$, then the polar form of $z$ is $r(\cos \theta  + i\sin \theta )$, where $r = |z| = \sqrt {{a^2} + {b^2}}$ and $\tan \theta  = \dfrac{b}{a} \cdot $ Here, $\theta$ is argument or amplitude of $z$ i.e., $\arg (z) = \theta$. The principal argument is a unique value of $\theta$ such that $- \pi  \leqslant \theta  \leqslant \pi$.

${\text{Given that,}}$

  $z = {\left( {1 + i\sqrt 3 } \right)^2}$

 $\Rightarrow z = 1 - 3 + 2i\sqrt 3  =  - 2 + i2\sqrt 3$

 $\Rightarrow \tan \alpha= \left| {\dfrac{{2\sqrt 3 }}{{ - 2}}} \right| = \left| { - \sqrt 3 } \right| = \sqrt 3 \left[ {\because \tan \alpha= \left| {\dfrac{{\operatorname{Im} \left( z \right)}}{{\operatorname{Re} \left( z \right)}}} \right|} \right]$

 $\Rightarrow \tan \alpha  = \tan \dfrac{\pi }{3}$

 $\Rightarrow \alpha  = \dfrac{\pi }{3}$ 

  $\because \operatorname{Re} \left( z \right) < 0{\text{and Im}}\left( z \right) > 0$

 $\Rightarrow \arg \left( z \right) = \pi  - \dfrac{\pi }{3} = \dfrac{{2\pi }}{3}$ 

34. Where does $z$ lie, if $\left| {\dfrac{{z - 5i}}{{z + 5i}}} \right| = 1$.

Ans: Given, $\left| {\dfrac{{z - 5i}}{{z + 5i}}} \right| = 1$

If${z_1} = {x_1} + i{y_1}$ and ${z_2} = {x_2} + i{y_2}$ then $\left| {{z_1}} \right| = \sqrt {x_1^2 + y_1^2}$ and $\left| {{z_2}} \right| = \sqrt {x_2^2 + y_2^2}$.

Also, use the modulus property i.e., $\left| {\dfrac{{{z_1}}}{{{z_2}}}} \right| = \dfrac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}}$.

Let $ z = x + iy$

Given that, $ \left| {\dfrac{{z - 5i}}{{z + 5i}}} \right| = \left| {\dfrac{{x + iy - 5i}}{{x + iy + 5i}}} \right|$

$\Rightarrow  \left| {\dfrac{{z - 5i}}{{z + 5i}}} \right| = \dfrac{{|x + i(y - 5)|}}{{|x + i(y + 5)|}}\left[ {\because \left| {\dfrac{{z - 5i}}{{z + 5i}}} \right| = 1} \right]$

$\Rightarrow  \left| {\dfrac{{z - 5i}}{{z + 5i}}} \right| = \dfrac{{\sqrt {{x^2} + {{(y - 5)}^2}} }}{{\sqrt {{x^2} + {{(y + 5)}^2}} }}$

On squaring both sides, we get 

${x^2} + {(y - 5)^2} = {x^2} + {(y + 5)^2}$

$\Rightarrow   - 10y =  + 10y$

$\Rightarrow  20y = 0$

$\therefore  y = 0$

So, z lies on real axis.

Choose the correct answer from the given four options indicated against each of the Exercises from 35 to 50 (M.C.Q) 

35. $z = sinx + icos2x$ and $cosx - isin2x$ are conjugate to each other for

$(A)\left( {x = n\pi } \right)$

$(B)x = \left( {n + \dfrac{1}{2}} \right)\dfrac{\pi }{2}$

$(C) x = 0$

(D) No value of x

Ans: Option C is correct answer

$z = \sin x + i\cos 2x$and $\cos x - i\sin 2x$ are conjugate to each other.

If${z_1} = {x_1} + i{y_1}$ and ${z_2} = {x_2} + i{y_2}$ then $\left| {{z_1}} \right| = \sqrt {x_1^2 + y_1^2}$ and $\left| {{z_2}} \right| = \sqrt {x_2^2 + y_2^2}$.

Also, use the modulus property i.e., $\left| {\dfrac{{{z_1}}}{{{z_2}}}} \right| = \dfrac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}}$.

Let $ z = \sin x + i\cos 2x$                   …..(i)

and $\bar z = \sin x - i\cos 2x$                      …..(ii)

Given that,

$\bar z = \cos x - i\sin 2x$

$\therefore$$\sin x - i\cos 2x = \cos x - i\sin 2x$

$\Rightarrow  \sin x = \cos x$ and $\cos 2x = \sin 2x$

$\Rightarrow\tan x = 1$ and $\tan 2x = 1$

$\Rightarrow  \tan x = \tan \dfrac{\pi }{4}$ and $\tan 2x = \tan \dfrac{\pi }{4}$

$\Rightarrow  x = n\pi  + \dfrac{\pi }{4}$ and $2x = n\pi  + \dfrac{\pi }{4}$

 $\Rightarrow  2x - x = 0$

 $\Rightarrow x = 0$ 

36. The real value of $\alpha $ for which the expression $\dfrac{{1 - isin\alpha }}{{1 + 2isin\alpha }}$ is purely real is

$(A)\left( {n + 1} \right)\dfrac{\pi }{2}$

$(B)\left( {2n + 1} \right)\dfrac{\pi }{2}$

$(C)n\pi$

(D) None of these

Ans: Option C is correct answer

$\dfrac{{1 - i\sin \alpha }}{{1 + 2i\sin \alpha }}$

First, convert the given expansion into $a + ib$ form and then check whether the complex number $a + ib$ is purely real.

Given expression, 

$z = \dfrac{{1 - i\sin \alpha }}{{1 + 2i\sin \alpha }}$

$= \dfrac{{(1 - i\sin \alpha )(1 - 2i\sin \alpha )}}{{(1 + 2i\sin \alpha )(1 - 2i\sin \alpha )}}$

 $\Rightarrow \dfrac{{1 - i\sin \alpha  - 2i\sin \alpha  + 2{i^2}{{\sin }^2}\alpha }}{{1 - 4{i^2}{{\sin }^2}\alpha }}$

 $\Rightarrow \dfrac{{1 - 3i\sin \alpha  - 2{{\sin }^2}\alpha }}{{1 + 4{{\sin }^2}\alpha }}$

 $\Rightarrow \dfrac{{1 - 2{{\sin }^2}\alpha }}{{1 + 4{{\sin }^2}\alpha }} - \dfrac{{3i\sin \alpha }}{{1 + 4{{\sin }^2}\alpha }}$ 

It is given that $z$ is a purely real.

  ${\therefore  \dfrac{{ - 3\sin \alpha }}{{1 + 4{{\sin }^2}\alpha }} = 0}$ 

 $\Rightarrow   - 3\sin \alpha  = 0$

 $\Rightarrow \sin \alpha  = 0$ 

  ${\alpha  = n\pi }$ 

37. If $z = x + iy$ lies in the third quadrant, then $\dfrac{{\bar z}}{z}$ also lies in the third quadrant, if

$(A)x > y > 0$

$(B)x < y < 0$

$(C)y < x < 0$

$(D)y > x > 0$

Ans: Option B is correct answer

Given, $z = x + iy$ lies in third quadrant. $x < 0$ and $y < 0$.

If $z = x + iy$ then $\bar z = x - iy$.

Given that, $z = x + iy$ lies in third quadrant. $x < 0$ and $y < 0$.

Now,

$\dfrac{{\bar z}}{z} = \dfrac{{x - iy}}{{x + iy}}$

 $\Rightarrow \dfrac{{(x - iy)(x - iy)}}{{(x + iy)(x - iy)}} = \dfrac{{{x^2} - {y^2} - 2ixy}}{{{x^2} + {y^2}}}$

${\text{ and }}$

$\dfrac{{\bar z}}{z} = \dfrac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} - \dfrac{{2ixy}}{{{x^2} + {y^2}}}$ 

Since, $\dfrac{{\bar z}}{z}$ also lies in the third quadrant.

$\therefore \dfrac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} < 0{\text{ and }}\dfrac{{ - 2xy}}{{{x^2} + {y^2}}} < 0$

${x^2} - {y^2} < 0{\text{ and }} - 2xy < 0$

${x^2} < {y^2}{\text{ and }}xy > 0$ 

So,

$x < y < 0$

38. The value of $(z + 3)(\bar z + 3)$ is equivalent to

$(A){\left| {z + 3} \right|^2}$

$(B)\left| {z - 3} \right|$

$(C){z^2} + 3$

(D) None of these

Ans: Option A is correct answer

Given, $(z + 3)(\bar z + 3)$

If $z = x + iy$ then $\bar z = x - iy$.

Given that, $(z + 3)(\bar z + 3)$

Let 

  $z = x + iy$

 $\Rightarrow (z + 3)(\bar z + 3) = (x + iy + 3)(x + 3 - iy)$

 $\Rightarrow {(x + 3)^2} - {(iy)^2} = {(x + 3)^2} + {y^2}$

 $\Rightarrow |x + 3 + iy{|^2} = |z + 3{|^2}$ 

39. If ${\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^x} = 1$, then

$(A)x = 2n + 1$

$(B)x = 4n$

$(C)x = 2n$

$(D)x = 4n + 1$ 

Ans: Option B is correct answer

${\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^x} = 1$

${i^{4n}} = 1,n \in N$

Given that, 

$ {\left( {\dfrac{{1 + i}}{{1 - i}}} \right)^x} = 1$

 $\Rightarrow  {\left[ {\dfrac{{(1 + i)(1 + i)}}{{(1 - i)(1 + i)}}} \right]^x} = 1$

 $\Rightarrow {\left[ {\dfrac{{1 + 2i + {i^2}}}{{1 - {i^2}}}} \right]^x} = 1$ 

 $\Rightarrow  {\left[ {\dfrac{{2i}}{{1 + 1}}} \right]^x} = 1$

 $\Rightarrow {\left[ {\dfrac{{2i}}{2}} \right]^x} = 1$ 

 $\Rightarrow  {i^x} = 1$

 $\Rightarrow {i^x} = \left( {{i^{4n}}} \right)\left[ {\because {i^{4n}} = 1,n \in N} \right]$ 

$\Rightarrow  x = 4n$

40. A real value of $x$ satisfies the equation $\left( {\dfrac{{3 - 4ix}}{{3 + 4ix}}} \right) = \alpha  - i\beta (\alpha ,\beta \in R)$, if ${\alpha ^2} + {\beta ^2}$ is equal to

(A)1

(B)-1

(C)2

(D)-2

Ans: Option A is correct answer

$\left( {\dfrac{{3 - 4ix}}{{3 + 4ix}}} \right) = \alpha  - i\beta (\alpha ,\beta  \in R)$

If $z = x + iy$ then $\bar z = x - iy$.

$\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$

Given equation, 

$ \left( {\dfrac{{3 - 4ix}}{{3 + 4ix}}} \right) = \alpha  - i\beta (\alpha ,\beta  \in R)$

$\Rightarrow  \left[ {\dfrac{{3 - 4ix}}{{3 + 4ix}}} \right] = \alpha  - i\beta$

Now, $ (\alpha  - i\beta ) = \dfrac{{(3 - 4ix)(3 - 4ix)}}{{(3 + 4ix)(3 - 4ix)}} = \dfrac{{9 + 16{i^2}{x^2} - 24ix}}{{9 - 16{i^2}{x^2}}}$

$\Rightarrow  \alpha  - i\beta  = \dfrac{{9 - 16{x^2} - 24ix}}{{9 + 16{x^2}}}$

$\Rightarrow  \alpha  - i\beta  = \dfrac{{9 - 16{x^2}}}{{9 + 16{x^2}}} - \dfrac{{i24x}}{{9 + 16{x^2}}}$                               …..(i)

$\therefore$$\alpha  + i\beta  = \dfrac{{9 - 16{x^2}}}{{9 + 16{x^2}}} + \dfrac{{i24x}}{{9 + 16{x^2}}}$                                   …..(ii)

${\text{So, }} (\alpha  - i\beta )(\alpha  + i\beta ) = {\left( {\dfrac{{9 - 16{x^2}}}{{9 + 16{x^2}}}} \right)^2} - {\left( {\dfrac{{i24x}}{{9 + 16{x^2}}}} \right)^2}$

$\therefore  {\alpha ^2} + {\beta ^2} = \dfrac{{81 + 256{x^4} - 288{x^2} + 576{x^2}}}{{{{\left( {9 + 16{x^2}} \right)}^2}}}$ 

 $\Rightarrow \dfrac{{81 + 256{x^4} + 288{x^2}}}{{{{\left( {9 + 16{x^2}} \right)}^2}}}$

 $\Rightarrow \dfrac{{{{\left( {9 + 16{x^2}} \right)}^2}}}{{{{\left( {9 + 16{x^2}} \right)}^2}}} = 1$ 

41. Which of the following is correct for any two complex numbers ${z_1}\text{ and }{z_2}$?

$(A)\left| {{z_1}{z_2}} \right| = \left| {{z_1}} \right|\left| {{z_2}} \right|$

$(B)\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$

$(C)arg\left( {{z_1}{z_2}} \right) = arg\left( {{z_1}} \right).arg\left( {{z_2}} \right)$

Ans: Option A is correct answer

complex numbers ${z_1}{\text{ and }}{z_2}$.

Let ${z_1} = {r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)$ and ${z_2} = {r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right)$ are two complex numbers.

  ${{\text{ Let }}} {{z_1} = {r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)}$ 

 $\Rightarrow  {\left| {{z_1}} \right| = {r_1}.....\left( i \right)}$ 

and $ {z_2} = {r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right)$

$\Rightarrow  \left| {{z_2}} \right| = {r_2}.....\left( {ii} \right)$

Now, 

  ${z_1}{z_2} = {r_1}{r_2}\left[ {\cos {\theta _1}\cos {\theta _2} + i\sin {\theta _1}\cos {\theta _2} + i\cos {\theta _1}\sin {\theta _2} + {i^2}\sin {\theta _1}\sin {\theta _2}} \right]$

$= {r_1}{r_2}\left[ {\cos \left( {{\theta _1} + {\theta _2}} \right) + i\sin \left( {{\theta _1} + {\theta _2}} \right)} \right]$

 $\Rightarrow \left| {{z_1}{z_2}} \right| = {r_1}{r_2}$

$\therefore \left| {{z_1}{z_2}} \right| = \left| {{z_1}} \right|\left| {{z_2}} \right|\left[ {{\text{Using equation }}\left( i \right){\text{ and }}\left( {ii} \right)} \right]$ 

42. The point represented by the complex number $(2 - i)$ is rotated about

origin through an angle $\dfrac{\pi }{2}$ in the clockwise direction, the new position of point is

$(A)1 + 2i$

$(B) - 1 - 2i$

$(C)2 + i$

$(D) - 1 + 2i$

Ans: Option B is correct answer

complex number $(2 - i)$

Here, $z < i\alpha $ is a complex number, where modulus is $r$ and argument $(\theta  + \alpha )$ If $P(z)$ rotates in clockwise sense through an angle $\alpha $, then its new position will be $z(\theta  - i\alpha )$

Given that, $z = 2 - i$

It is rotated about origin through an angle $\dfrac{\pi }{2}$ in the clockwise direction.

$\therefore  {\text{ New position}} = z{e^{ - i\pi /2}}$

 $\Rightarrow (2 - i){e^{ - i\pi /2}} = (2 - i)\left[ {\cos \left( {\dfrac{{ - \pi }}{2}} \right) + i\sin \left( {\dfrac{{ - \pi }}{2}} \right)} \right]$

 $\Rightarrow (2 - i)[0 - i] =  - 2i - 1 =  - 1 - 2i$ 

43. If $x,y\in R$, then $x + iy$ is a non-real complex number, if

(A) $x = 0$

(B) $y = 0$

(C) $y =0$

(D) $y \ne 0$

Ans: Option D is correct answer

Given, $x,y \in R$

$x + iy$ is non-real complex number if and only if $y \ne 0$.

Given that,

$x,y \in R$

Then, $x + iy$ is non-real complex number if and only if $y \ne 0$.

44. If $a + ib = c + id$, if

(A) ${a^2} + {c^2} = 0$

(B) ${b^2} + {c^2} = 0$

(C) ${b^2} + {d^2} = 0$

(D) ${a^2} + {b^2} = {c^2} + {d^2}$

Ans: Option D is correct answer

$a + ib = c + id$

If two complex numbers ${z_1} = {x_1} + i{y_1}$ and ${z_2} = {x_2} + i{y_2}$ are equal, then

$\left| {{z_1}} \right| = \left| {{z_2}} \right|$

 $\Rightarrow \sqrt {x_1^2 + y_1^2}  = \sqrt {x_2^2 + y_2^2}$ 

Given that, $a + ib = c + id$

$\Rightarrow  |a + ib| = |c + id|$

$\Rightarrow  \sqrt {{a^2} + {b^2}}  = \sqrt {{c^2} + {d^2}}$

On squaring both sides, we get 

${a^2} + {b^2} = {c^2} + {d^2}$

45. The complex number $z$ which satisfies the condition $\left| {\dfrac{{i + z}}{{i - z}}} \right| = 1$ lies on

(A) circle ${x^2} + {y^2} = 1$

(B) the $x$- axis

(C) the $y$- axis

(D) the line $x + y = 1$

Ans: Option B is correct answer

$\left| {\dfrac{{i + z}}{{i - z}}} \right| = 1$

Let $z$ be a complex number and $z = x + iy$ then $\left| z \right| = \sqrt {{x^2} + {y^2}}$.

Given that, $\left| {\dfrac{{i + z}}{{i - z}}} \right| = 1$

Let $ z = x + iy$

$\therefore  \left| {\dfrac{{x + i(y + 1)}}{{ - x - i(y - 1)}}} \right| = 1$

 $\Rightarrow \dfrac{{{x^2} + {{(y + 1)}^2}}}{{{x^2} + {{(y - 1)}^2}}} = 1$ 

$\Rightarrow  {x^2} + {(y + 1)^2} = {x^2} + {(y - 1)^2}$

 $\Rightarrow  4y = 0$

 $\Rightarrow y = 0$ 

So, $z$ lies on $x$-axis (real axis).

46. If $z$ is a complex number, then

(A) $\left| {{z^2}} \right| > |z|$

(B) $\left| {{z^2}} \right| = |z{|^2}$

(C) $\left| {{z^2}} \right| < |z{|^2}$

(D) $\left| {{z^2}} \right| \geq  |z{|^2}$

Ans: Option B is correct answer

$z$ is a complex number.

If $z$ is a complex number, then $z = x + iy$ $|z| = |x + |y\mid$ and $|z{|^2} = |x + iy{|^2}$

If $z$ is a complex number, then $z = x + iy$ $|z| = |x + |y\mid$ and $|z{|^2} = |x + iy{|^2}$

$\Rightarrow  |z{|^2} = {x^2} + {y^2}.....\left( i \right)$

and $ {z^2} = {(x + iy)^2} = {x^2} + {i^2}{y^2} + i2xy$

${z^2} = {x^2} - {y^2} + i2xy$

$\Rightarrow \left| {{z^2}} \right| = \sqrt {{{\left( {{x^2} - {y^2}} \right)}^2} + {{(2xy)}^2}}$

$\Rightarrow  \left| {{z^2}} \right| = \sqrt {{x^4} + {y^4} - 2{x^2}{y^2} + 4{x^2}{y^2}}$

$\Rightarrow  \left| {{z^2}} \right| = \sqrt {{x^4} + {y^4} + 2{x^2}{y^2}}  = \sqrt {{{\left( {{x^2} + {y^2}} \right)}^2}}$

$\Rightarrow  \left| {{z^2}} \right| = {x^2} + {y^2}.....\left( {ii} \right)$

From Eqs. (i) and (ii), ${\left| {z} \right|^2} = \left| {{z^2}} \right|$

47. $\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$ is possible, if

(A) ${z_2} = {\bar z_1}$

(B) ${z_2} = \dfrac{1}{{{z_1}}}$

(C) $arg\left( {{z_1}} \right) = arg\left( {{z_2}} \right)$

(D) $\left| {{z_1}} \right| = \left| {{z_2}} \right|$

Ans: Option C is correct answer

 $\left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$

Let $z$ be a complex number and $z = x + iy$ then $\left| z \right| = \sqrt {{x^2} + {y^2}}$.

Given that, $ \left| {{z_1} + {z_2}} \right| = \left| {{z_1}} \right| + \left| {{z_2}} \right|$

$\Rightarrow \left| {{r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right) + {r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right)} \right| = \left| {{r_1}\left( {\cos {\theta _1} + i\sin {\theta _1}} \right)} \right|$

$+ \left| {{r_2}\left( {\cos {\theta _2} + i\sin {\theta _2}} \right)} \right|$

$\Rightarrow \left| {\left( {{r_1}\cos {\theta _1} + {r_2}\cos \theta } \right) + i\left( {{r_1}\sin {\theta _1} + {r_2}\sin {\theta _2}} \right)} \right| = {r_1} + {r_2}$

$\Rightarrow  \sqrt {r_1^2{{\cos }^2}{\theta _1} + r_2^2{{\cos }^2}{\theta _2} + 2{r_1}{r_2}\cos {\theta _1}\cos {\theta _2} + r_1^2{{\sin }^2}{\theta _1} + r_2^2{{\sin }^2}{\theta _2}}$

$\sqrt { + 2{r_1}{r_2}\sin {\theta _1}\sin {\theta _2}}  = {r_1} + {r_2}$

$\Rightarrow \sqrt {r_1^2 + r_2^2 + 2{r_1}{r_2}\left[ {\cos \left( {{\theta _1} - {\theta _2}} \right)} \right]}  = {r_1} + {r_2}$

On squaring both sides, we get 

$r_1^2 + r_2^2 + 2{r_1}{r_2}\cos \left( {{\theta _1} - {\theta _2}} \right) = r_1^2 + r_2^2 + 2{r_1}{r_2}$

$\Rightarrow  2{r_1}{r_2}\left[ {1 - \cos \left( {{\theta _1} - {\theta _2}} \right)} \right] = 0$

$\Rightarrow1 - \cos \left( {{\theta _1} - {\theta _2}} \right) = 0$

$\Rightarrow\cos \left( {{\theta _1} - {\theta _2}} \right) = 1$

$\Rightarrow \cos \left( {{\theta _1} - {\theta _2}} \right) = \cos {0^\circ }$

$\Rightarrow  {\theta _1} - {\theta _2} = {0^\circ }$

$\Rightarrow{\theta _1} = {\theta _2}$

$\therefore$$\arg \left( {{z_1}} \right) = \arg \left( {{z_2}} \right)$

48. The real value of $\theta$ for which the expression $\dfrac{{1 + icos\theta }}{{1 - 2icos\theta }}$ is a real number is

(A) $n\pi  + \dfrac{\pi }{4}$

(B) $n\pi  + {( - 1)^n}\dfrac{\pi }{4}$

(C) $2n\pi  \pm \dfrac{\pi }{2}$

(D) None of these

Ans: Option C is correct answer

$\dfrac{{1 + i\cos \theta }}{{1 - 2i\cos \theta }}$

Simplify the given equation and use the formula $\cos \dfrac{\pi }{2} = 0$.

Given expression,

$= \dfrac{{1 + i\cos \theta }}{{1 - 2i\cos \theta }} = \dfrac{{(1 + i\cos \theta )(1 + 2i\cos \theta )}}{{(1 - 2i\cos \theta )(1 + 2i\cos \theta )}}$

 $\Rightarrow \dfrac{{1 + i\cos \theta  + 2i\cos \theta  + 2{i^2}{{\cos }^2}\theta }}{{1 - 4{i^2}{{\cos }^2}\theta }}$

 $\Rightarrow \dfrac{{1 + 3i\cos \theta  - 2{{\cos }^2}\theta }}{{1 + 4{{\cos }^2}\theta }}$ 

For real value of $\theta ,$

$\dfrac{{3\cos \theta }}{{1 + 4{{\cos }^2}\theta }} = 0$

 $\Rightarrow 3\cos \theta  = 0$

 $\Rightarrow \cos \theta  = \cos \dfrac{\pi }{2}$

 $\Rightarrow \theta  = 2n\pi  \pm \dfrac{\pi }{2}$ 

49. The value of arg $(x)$, when $x < 0$ is

(A) 0

(B) $\dfrac{\pi }{2}$

(C) $\pi$

(D) None of these

Ans: Option C is correct answer

$x < 0$

Let $z$be a complex number and $z = x + iy$ then $\left| z \right| = \sqrt {{x^2} + {y^2}}$.

Let

  $z = x + 0i{\text{ and }}x < 0$

  $|z| = \sqrt {{{( - 1)}^2} + \left( {{0^2}} \right)}  = 1$ 

Since, the point $(x, 0)$ represent $z = x + 0i$ lies on the negative side of real axis. 

$\therefore$ Principal $\arg (z) = \pi$

50. If $f(z) = \dfrac{{7 - z}}{{1 - {z^2}}}$, where $z = 1 + 2i$, then $|f(z)|$ is equal to

(A) $\dfrac{{|z|}}{2}$

(B) $\left| {z} \right|$

(C) $2\left| {z} \right|$

(D) None of these

Ans: Option A is correct answer

$f(z) = \dfrac{{7 - z}}{{1 - {z^2}}}$, where $z = 1 + 2i$

Let $z$ be a complex number and $z = x + iy$ then $\left| z \right| = \sqrt {{x^2} + {y^2}}$.

${{\text{ Let }}} {z = 1 + 2i}$ 

 $\Rightarrow {|z| = \sqrt {1 + 4}  = \sqrt 5 } $

Now, 

  $f(z) = \dfrac{{7 - z}}{{1 - {z^2}}} = \dfrac{{7 - 1 - 2i}}{{1 - {{(1 + 2i)}^2}}}$

 $\Rightarrow \dfrac{{6 - 2i}}{{1 - 1 - 4{i^2} - 4i}} = \dfrac{{6 - 2i}}{{4 - 4i}}$ 

$\Rightarrow \dfrac{{(3 - i)(2 + 2i)}}{{(2 - 2i)(2 + 2i)}} = \dfrac{{6 - 2i + 6i - 2{i^2}}}{{4 - 4{i^2}}}$

 $\Rightarrow \dfrac{{6 + 4i + 2}}{{4 + 4}} = \dfrac{{8 + 4i}}{8}$

 $\Rightarrow 1 + \dfrac{1}{2}i$ 

 $f(z) = 1 + \dfrac{1}{2}i$

$\therefore |f(z)| = \sqrt {1 + \dfrac{1}{4}}$ 

$\Rightarrow \sqrt {\dfrac{{4 + 1}}{4}}  = \dfrac{{\sqrt 5 }}{2} = \dfrac{{|z|}}{2}$

An overview of NCERT Exemplar for Class 11 Mathematics - Complex Numbers and Quadratic Equations

NCERT Exemplar for Class 11 Mathematics - Complex Numbers and Quadratic Equations - is a Chapter filled with equations and different concepts to cover. Vedantu’s team is aware that students alone might find it difficult to comprehend such equations in one attempt. So, to offer the best resolution, their team of expert tutors of the respective subject help and guide students in the right direction.

The entire syllabus has been planned and covered by Vedantu in such a manner that students not only study well, but they are also able to make the most of their study by planning everything in a calculative manner. Students feel encouraged to create a regular timetable so that they won’t miss a single question by the time the exam reaches closer. Vedantu even assures students of their ability to provide round the clock guidance whenever they find themselves stuck in tricky questions. This way, students can learn everything in the flow which contributes not only to their daily goals but it creates consistency so that they can reap the complete benefits of their persistent hard work later. 

Are you ready to boost your performance? Then, choose Vedantu today and secure your future study goals.