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

$(a)\;2\;[cos(\large\frac{\pi}{3})+i \;sin(\large\frac{\pi}{3})]\qquad(b)\;2\;[cos(\large\frac{\pi}{6})+i\; sin(\large\frac{\pi}{6})]\qquad(c)\;\sqrt{2}\;[cos (-\large\frac{\pi}{2})+i \;sin(-\large\frac{\pi}{2})]\qquad(d)\;2\;[cos(\large\frac{\pi}{2})+i \;sin(\large\frac{\pi}{2})]$

Answer : $\;2\;[cos(\large\frac{\pi}{6})+i\; sin(\large\frac{\pi}{6})]$
Explanation :
$\;\sqrt{3}+i\;$ Let $\;r cos \theta =\sqrt{3} \;$ and $\;r sin \theta = 1$
on squaring and adding , we obtain
$(r cos \theta)^{2}+(r sin \theta)^{2} = (\sqrt{3})^{2}+(1)^{2}$
$r^{2}(cos^{2} \theta + sin^{2} \theta) = 3+ 1$
$r^{2} = 4 \qquad [cos^{2} \theta + sin^{2} \theta=1]$
$r =\sqrt{4} =2\qquad [conventionally , r > 0 ]$
Therefore , $\; 2 cos \theta = \sqrt{3} \;$ and $\; 2 sin \theta = 1$
$cos \theta = \large\frac{\sqrt{3}}{2}\;$ and $\;sin \theta = \large\frac{1}{2}$
$\theta = \large\frac{\pi}{6}\qquad [\theta \;lies\; in\;the\;1^{st} \; quadrant]$
$\sqrt{3}+i = r cos \theta + i r sin \theta$
$=2\;[cos (\large\frac{\pi}{6})+i \;sin(\large\frac{\pi}{6})]\;$ this is the required polar form .