Question

Convert the following complex number in the polar form : $\sqrt{3}+i$

Answer 1

Polar form of $\sqrt{3}+i$

Step 1: Given $z=\sqrt{3}+i$

Step 2: Modulus of complex number
$r=\sqrt{x^2+y^2}$
$\Rightarrow r=\sqrt{(\sqrt{3}{)}^2+(1{)}^2}\Rightarrow r=2$
$\therefore \text{Modulus}r=2$

Step 3: $\because z$ lies in the first quadrant.
$\therefore \text{Argument}={\tan }^{-1}\left|\frac{y}{x}\right|={\tan }^{-1}\left|\frac{1}{\sqrt{3}}\right|=\frac{\pi }{6}$

$\therefore \text{Argument}=\frac{\pi }{6}$

Step 4 : Polar form : $z=r(\cos \theta +i\sin \theta )$
$z=2(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right))$
Polar form of complex number $\sqrt{3}+iis2(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right))$