Convert each of the complex numbers in the polar form: (5) -1-i

Convert each of the complex numbers in the polar form:

Q : 6 $-1-i$

Answers (1)

Given problem is
$z=-1-i$
Now, let
$r\cos \theta = -1 \ \ \ and \ \ \ r\sin \theta = -1$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (-1)^2+(-1)^2$ $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 1+1$
$r^2 =2$
$r= \sqrt2$ $(\because r > 0)$
Therefore, the modulus is $\sqrt2$
Now,
$\sqrt2\cos \theta = -1 \ \ \ and \ \ \ \sqrt2\sin \theta = -1$
$\cos \theta = -\frac{1}{\sqrt2} \ \ \ and \ \ \ \sin \theta =-\frac{1}{\sqrt2}$
Since values of both $\cos \theta$ and $\sin \theta$ is negative and we know that this is the case in III quadrant
Therefore,
$\theta =-\left ( \pi - \frac{\pi}{4} \right )= -\frac{3\pi}{4}\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ III \ quadrant)$
Therefore,
$-1-i= r\cos \theta +ir\sin \theta$
$= \sqrt2\cos \left ( -\frac{3\pi}{4} \right ) +i\sqrt2\sin \left (- \frac{3\pi}{4} \right )$
$= \sqrt2\left ( \cos \left ( -\frac{3\pi}{4} \right ) +i\sin \left ( -\frac{3\pi}{4} \right ) \right )$

Therefore, the required polar form is $\sqrt2\left ( \cos \left (- \frac{3\pi}{4} \right ) +i\sin \left (- \frac{3\pi}{4} \right ) \right )$