# Convert each of the complex numbers in the polar form: Q : 8    $i$

G Gautam harsolia

Given problem is
$z = i$
Now, let
$r\cos \theta = 0 \ \ \ and \ \ \ r\sin \theta = 1$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (0)^2+(1)^2$                                                     $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 0+1$
$r^2 =1$
$r= 1$                                                                                                                         $(\because r > 0)$
Therefore, the modulus is 1
Now,
$1\cos \theta =0 \ \ \ and \ \ \ 1\sin \theta = 1$
$\cos \theta =0\ \ \ and \ \ \ \sin \theta =1$
Since values of Both $\cos \theta$  and $\sin \theta$ is Positive  and  we know that this is the case in  I quadrant
Therefore,
$\theta =\frac{\pi}{2}\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ I \ quadrant)$
Therefore,
$i= r\cos \theta +ir\sin \theta$
$= 1\cos \left (\frac{\pi}{2} \right ) +i1\sin \left (\frac{\pi}{2} \right )$
$= \cos \frac{\pi}{2} +i\sin\frac{\pi}{2}$

Therefore, the required polar form is   $\cos \frac{\pi}{2} +i\sin\frac{\pi}{2}$

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