Question

Find the modulus and the argument of the complex number$z=-1-i\sqrt{3}$.

Answer 1

Hint-These types of questions can be solved by using the formula of modulus and argument of
 the complex number.
Given complex number is
$z=-1-i\sqrt{3}$
Now we know that the general form of complex number is
$z=x+iy$
Now comparing the above two we get,
$x=-1$ and $y=-\sqrt{3}$
Now let’s find the modulus of the complex number.
We know that the modulus of a complex number is $\left|z\right|$
$\left|z\right|=\sqrt{x^2+y^2}$
Now putting the value of $x$ and $y$ we get,
$$\begin{gathered} \left|z\right|=\sqrt{{(-1)}^2+{(-\sqrt{3})}^2} \\ \left|z\right|=\sqrt{1+3} \\ \left|z\right|=\sqrt{4} \\ \left|z\right|=2 \end{gathered}$$
Therefore, the modulus of a given complex number is $2$.
Now let’s find the argument of the complex number.
Now we know that the general form of complex number is
$z=x+iy$
Let $x$ be $r\cos \theta$ and$y$ be $r\sin \theta$ where $r$ is the modulus of the complex number.
Now putting the values of $x$ and $y$ in $z$ we get,
$z=r\cos \theta +ir\sin \theta$
Now comparing the above two we get,
$-1-i\sqrt{3}=r\cos \theta +ir\sin \theta$
Now, comparing the real parts we get,
$\text{ - 1 = }r\cos \theta$
Now, putting the value of $r$ in the above equation we get,
$$\begin{gathered} \text{ - 1 = 2}\cos \theta \\ \text{or }\cos \theta ={\displaystyle \frac{-1}{2}} \end{gathered}$$
Similarly, compare the imaginary parts and put the value of $r$ we get,
$$\begin{gathered} -\sqrt{3}=2\sin \theta \\ \text{or }\sin \theta =-{\displaystyle \frac{\sqrt{3}}{2}} \end{gathered}$$
Hence, $\sin \theta =-{\displaystyle \frac{\sqrt{3}}{2}}\text{ and }\cos \theta ={\displaystyle \frac{-1}{2}}$
or$\theta \text{ = }{60}^{\circ }$
Now we can clearly see that the values of both $\sin \theta$ and$\cos \theta$ are negative.
And we know that they both are negative in $3^{rd}$ quadrant.
Therefore, the argument is in $3^{rd}$ quadrant.
Argument$\text{ = - (18}{0}^{\circ }-\theta \text{)}$
$$\begin{gathered} \text{ = - (18}{0}^{\circ }-\theta \text{)} \\ =\text{ - (18}{0}^{\circ }-{60}^{\circ }\text{)} \\ \text{ = - (12}{\text{0}}^{\circ }\text{)} \\ \text{ = - 12}{0}^{\circ } \end{gathered}$$
Now converting it in $\pi$ form we get,
$$\begin{gathered} =-{120}^{\circ }\times {\displaystyle \frac{\pi }{{180}^{\circ }}} \\ =-{\displaystyle \frac{2\pi }{3}} \end{gathered}$$
Hence, the argument of complex number is $-{\displaystyle \frac{2\pi }{3}}$
Note- Whenever we face such types of questions the key concept is that we simply compare the given
 complex number with its general form and then find the value of $x$ and $y$ and put it in the formula
 of modulus and argument of the complex number