Convert the complex number frac{-16}{1+i sqrt | vedclass

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(A) The given complex number is $z = \frac{-16}{1+i \sqrt{3}}$.Multiply the numerator and denominator by the conjugate $1-i \sqrt{3}$:$z = \frac{-16(1

Vedclass · Accepted Answer

$8\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)$

Vedclass · Answer

(A) The given complex number is $z = \frac{-16}{1+i \sqrt{3}}$.Multiply the numerator and denominator by the conjugate $1-i \sqrt{3}$:$z = \frac{-16(1-i \sqrt{3})}{(1+i \sqrt{3})(1-i \sqrt{3})} = \frac{-16(1-i \sqrt{3})}{1^2 - (i \sqrt{3})^2} = \frac{-16(1-i \sqrt{3})}{1+3} = \frac{-16(1-i \sqrt{3})}{4} = -4 + i 4 \sqrt{3}$.Let $z = r(\cos 	heta + i \sin 	heta)$,where $r = \sqrt{(-4)^2 + (4 \sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8$.Then $\cos 	heta = \frac{-4}{8} = -\frac{1}{2}$ and $\sin 	heta = \frac{4 \sqrt{3}}{8} = \frac{\sqrt{3}}{2}$.Since $\cos 	heta  0$,$	heta$ lies in the second quadrant.$	heta = \pi - \frac{\pi}{3} = \frac{2 \pi}{3}$.Thus,the polar form is $8\left(\cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3}ight)$.

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

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