# Convert the complex number $\;\large\frac{-16}{1+i \sqrt{3}} \;$ into polar form

$(a)\;8(cos \large\frac{2\pi}{3}+ i sin\large\frac{2\pi}{3})\qquad(b)\;8(cos \large\frac{\pi}{5}+ i sin\large\frac{\pi}{5})\qquad(c)\;(cos \large\frac{2\pi}{3}+ i sin\large\frac{2\pi}{3})\qquad(d)\;0$

Answer : $\;8(cos \large\frac{2\pi}{3}+ i sin\large\frac{2\pi}{3})$
Explanation :
the complex given number $\;\large\frac{-16}{1+i \sqrt{3}} \;$
$\large\frac{-16}{1+i \sqrt{3}} = \large\frac{-16}{1+i \sqrt{3}} \times \large\frac{1-i \sqrt{3}}{1-i \sqrt{3}}$
$= \large\frac{-16( 1-i\sqrt{}3 )}{1^{2} - (i \sqrt{3})^{2}}$
$= \large\frac{-16( 1-i\sqrt{}3 )}{4}= -4(1 - i \sqrt{3})$
$= -4 +i 4 \sqrt{3}$
Let $\; r cos \theta = -4 \;$ and $\; r sin \theta = 4 \sqrt{3}$
By squaring and adding , we get
$r^{2} (cos^{2} \theta + sin^{2} \theta)^{2} = 16 + 48$
$r = \sqrt{64} = 8 \qquad$ [conventionally , r > 0]
Therefore , $\; cos \theta =- \large\frac{1}{2}\; , sin \theta =\large\frac{\sqrt{3}}{2}\;$
$\theta =\pi - \large\frac{\pi}{3}= \large\frac{2 \pi}{3}$
Therefore , required polar form is $\;8(cos \large\frac{2\pi}{3}+ i sin\large\frac{2\pi}{3})$