# Convert each of the complex numbers in the polar form: Q: 6        $-3$

G Gautam harsolia

Given problem is
$z=-3$
Now, let
$r\cos \theta = -3 \ \ \ and \ \ \ r\sin \theta = 0$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (-3)^2+(0)^2$                                                     $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 9+0$
$r^2 =9$
$r= 3$                                                                                                                         $(\because r > 0)$
Therefore, the modulus is 3
Now,
$3\cos \theta =- 3 \ \ \ and \ \ \ 3\sin \theta = 0$
$\cos \theta = -1\ \ \ and \ \ \ \sin \theta =0$
Since values of  $\cos \theta$ is negative and $\sin \theta$ is Positive  and  we know that this is the case in  II quadrant
Therefore,
$\theta =\pi\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ II \ quadrant)$
Therefore,
$-3= r\cos \theta +ir\sin \theta$
$= 3\cos \left (\pi \right ) +i3\sin \left (\pi \right )$
$= 3\left ( \cos \pi +i\sin\pi \right )$

Therefore, the required polar form is   $3\left ( \cos \pi +i\sin\pi \right )$

Exams
Articles
Questions