Answer
Verified
472.2k+ views
Hint: A complex number is a number generally represented as\[z = a + ib\], where \[a\] and \[b\] is a real number represented on the real axis whereas \[i\] is an imaginary unit represented on the imaginary axis whose value is \[i = \sqrt { - 1} \]. Modulus of a complex number is the length of a line segment on a real and imaginary axis generally denoted by $|z| = \sqrt {{a^2} + {b^2}} $. Conjugate of a complex number is the negation of the imaginary part of the complex number while keeping the real part as it is such as for \[z = a + ib\], the complex conjugate will be defined as \[z = a - ib\].
In this question, two complex numbers are given, and we need to determine the relation between the conjugate of one complex number with the original form of another complex number for which we need to follow the concept of argument and properties of the complex conjugate of the numbers.
Complete step by step solution: Rearranging the given equation $|z{|^2}w - |w{|^2}z = z - w{\text{ and, (}}z \ne w{\text{)}}$ as:
\[
|z{|^2}w - |w{|^2}z = z - w{\text{ }} \\
|z{|^2}w + w = z + |w{|^2}z \\
w\left( {|z{|^2} + 1} \right) = z\left( {|w{|^2} + 1} \right) \\
\dfrac{w}{z} = \left( {\dfrac{{|w{|^2} + 1}}{{|z{|^2} + 1}}} \right) - - - - (i) \\
\]
As equation (i), does not have any imaginary part so, from equation (i), it can also be written as:
\[\dfrac{{\bar w}}{{\bar z}} = \left( {\dfrac{{|w{|^2} + 1}}{{|z{|^2} + 1}}} \right) - - - - (ii)\]
Dividing equation (i) and equation (ii) we get,
\[
\dfrac{{\left( {\dfrac{w}{z}} \right)}}{{\left( {\dfrac{{\bar w}}{{\bar z}}} \right)}} = \dfrac{{\left( {\dfrac{{|w{|^2} + 1}}{{|z{|^2} + 1}}} \right)}}{{\left( {\dfrac{{|w{|^2} + 1}}{{|z{|^2} + 1}}} \right)}} \\
\dfrac{{w\bar z}}{{\bar wz}} = 1 \\
w\bar z = \bar wz = 1 \\
\]
Hence, for two complex numbers $z{\text{ and }}w$, such that $|z{|^2}w - |w{|^2}z = z - w{\text{ and, (}}z \ne w{\text{)}}$ then $z\bar w = 1$.
Option B is correct.
Note: It is interesting to note here that, the complex conjugate of a real number is the number itself as it does not contain any imaginary term with it such as for $z = pq$, the complex conjugate will be the number itself i.e., $z = pq$ as it does not have any imaginary part for the negation.
In this question, two complex numbers are given, and we need to determine the relation between the conjugate of one complex number with the original form of another complex number for which we need to follow the concept of argument and properties of the complex conjugate of the numbers.
Complete step by step solution: Rearranging the given equation $|z{|^2}w - |w{|^2}z = z - w{\text{ and, (}}z \ne w{\text{)}}$ as:
\[
|z{|^2}w - |w{|^2}z = z - w{\text{ }} \\
|z{|^2}w + w = z + |w{|^2}z \\
w\left( {|z{|^2} + 1} \right) = z\left( {|w{|^2} + 1} \right) \\
\dfrac{w}{z} = \left( {\dfrac{{|w{|^2} + 1}}{{|z{|^2} + 1}}} \right) - - - - (i) \\
\]
As equation (i), does not have any imaginary part so, from equation (i), it can also be written as:
\[\dfrac{{\bar w}}{{\bar z}} = \left( {\dfrac{{|w{|^2} + 1}}{{|z{|^2} + 1}}} \right) - - - - (ii)\]
Dividing equation (i) and equation (ii) we get,
\[
\dfrac{{\left( {\dfrac{w}{z}} \right)}}{{\left( {\dfrac{{\bar w}}{{\bar z}}} \right)}} = \dfrac{{\left( {\dfrac{{|w{|^2} + 1}}{{|z{|^2} + 1}}} \right)}}{{\left( {\dfrac{{|w{|^2} + 1}}{{|z{|^2} + 1}}} \right)}} \\
\dfrac{{w\bar z}}{{\bar wz}} = 1 \\
w\bar z = \bar wz = 1 \\
\]
Hence, for two complex numbers $z{\text{ and }}w$, such that $|z{|^2}w - |w{|^2}z = z - w{\text{ and, (}}z \ne w{\text{)}}$ then $z\bar w = 1$.
Option B is correct.
Note: It is interesting to note here that, the complex conjugate of a real number is the number itself as it does not contain any imaginary term with it such as for $z = pq$, the complex conjugate will be the number itself i.e., $z = pq$ as it does not have any imaginary part for the negation.
Recently Updated Pages
A ray of light passes through an equilateral prism class 12 physics JEE_Main
The size of the image of an object which is at infinity class 12 physics JEE_Main
When a glass slab is placed on a cross made on a sheet class 12 physics JEE_Main
Rays from Sun converge at a point 15 cm in front of class 12 physics JEE_Main
For the circuit shown in figure the equivalent capacitance class 12 physics JEE_Main
If on applying the potential of 20 V on a conductor class 12 physics JEE_Main
Trending doubts
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
What are the major means of transport Explain each class 12 social science CBSE
Explain sex determination in humans with the help of class 12 biology CBSE
How much time does it take to bleed after eating p class 12 biology CBSE