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# Convert each of the complex numbers in the polar form: (7) sqrt 3 + i

Convert each of the complex numbers in the polar form:

Q : 7    $\sqrt{3}+i$

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Given problem is
$z=\sqrt3+i$
Now, let
$r\cos \theta = \sqrt3 \ \ \ and \ \ \ r\sin \theta = 1$
Square and add both the sides
$r^2(\cos^2\theta +\sin^2\theta)= (\sqrt3)^2+(1)^2$                                                     $(\because \cos^2\theta +\sin^2\theta = 1)$
$r^2= 3+1$
$r^2 =4$
$r= 2$                                                                                                                         $(\because r > 0)$
Therefore, the modulus is 2
Now,
$2\cos \theta =\sqrt3 \ \ \ and \ \ \ 2\sin \theta = 1$
$\cos \theta = \frac{\sqrt3}{2}\ \ \ and \ \ \ \sin \theta =\frac{1}{2}$
Since values of Both $\cos \theta$  and $\sin \theta$ is Positive  and  we know that this is the case in  I quadrant
Therefore,
$\theta =\frac{\pi}{6}\ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ I \ quadrant)$
Therefore,
$\sqrt3+i= r\cos \theta +ir\sin \theta$
$= 2\cos \left (\frac{\pi}{6} \right ) +i2\sin \left (\frac{\pi}{6} \right )$
$= 2\left ( \cos \frac{\pi}{6} +i\sin\frac{\pi}{6} \right )$

Therefore, the required polar form is   $2\left ( \cos \frac{\pi}{6} +i\sin\frac{\pi}{6} \right )$

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