Question

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

Answer 1

$z=\sqrt{3}+i$
Let $rcos\theta =\sqrt{3}andrsin\theta =1$
On squaring and adding, we obtain
$r^{2}co{s}^2\theta +r^{2}si{n}^2\theta =(\sqrt{3}{)}^2+(1{)}^2\Rightarrow r^2(co{s}^2\theta +si{n}^2\theta )=3+1\Rightarrow r^2=4\Rightarrow r=\sqrt{4}=2[Conventionally,r>0]\therefore 2cos\theta =\sqrt{3}and2sin\theta =1\Rightarrow cos\theta =\frac{\sqrt{3}}{2}andsin\theta =\frac{1}{2}\therefore \theta =\frac{\pi }{6}\left[As\theta liesintheIquadrant\right]\therefore \sqrt{3}+i=rcos\theta +irsin\theta =2cos\frac{\pi }{6}+i2sin\frac{\pi }{6}=2(cos\frac{\pi }{6}+isin\frac{\pi }{6})$
This is the required polar form.