Question

Convert $\frac{-16}{1+i\sqrt{3}}$ into polar form.

• A. $8\left(cos\frac{2\pi }{3}+isin\frac{2\pi }{3}\right)$
• B. $4\left(cos\frac{\pi }{3}+isin\frac{\pi }{3}\right)$
• C. $cos\frac{2\pi }{3}+isin\frac{2\pi }{3}$
• D. None of these

Answer 1

Answer: (A)

Let $z=\frac{-16}{1+i\sqrt{3}}$
$=\frac{-16}{1+i\sqrt{3}}\times \frac{1-i\sqrt{3}}{1-i\sqrt{3}}$
$=\frac{-16\left(1-i\sqrt{3}\right)}{1+3}$
$=-4\left(1-i\sqrt{3}\right)$
$=-4+i4\sqrt{3}$
Let $-4=rcos\theta ,4\sqrt{3}=rsin\theta$
By squaring and adding, we get
$16+48=r^2\left(co{s}^2\theta +si{n}^2\theta \right)$
$\Rightarrow r^2=64$
$\Rightarrow r=8$
Hence, cos $\theta =\frac{-1}{2}$,
sin $\theta =\frac{\sqrt{3}}{2}$
$\Rightarrow \theta =\pi -\frac{\pi }{3}=\frac{2\pi }{3}$
Thus, the required polar form is $8\left(cos\frac{2\pi }{3}+isin\frac{2\pi }{3}\right)$