If z= 1+i 3 3+i , then (z)^ 100 lies in

MCQ

If $z=\frac{1+i \sqrt{3}}{\sqrt{3}+i}$, then $(z)^{100}$ lies in

A
$I^{\text {st }}$ quadrant
B
$II^{\text {nd }}$ quadrant
✓
III ${ }^{\text {rd }}$ quadrant
D
IV ${ }^{\text {th }}$ quadrant

✓

Answer

Correct option: C.

III ${ }^{\text {rd }}$ quadrant

(C)
$z=\frac{1+i \sqrt{3}}{\sqrt{3}+i} \Rightarrow z=\frac{1+i \sqrt{3}}{\sqrt{3}+i} \times \frac{\sqrt{3}-i}{\sqrt{3}-i}$
$z=\frac{\sqrt{3}+3 i-i+\sqrt{3}}{3+1}=\frac{2(\sqrt{3}+i)}{4}$
$\Rightarrow \quad z=\frac{\sqrt{3}+i}{2}=\left[\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right]$
Now $\bar{z}=\cos \frac{\pi}{6}-i \sin \frac{\pi}{6}$
$\Rightarrow(\bar{z})^{100}=\left[\cos \frac{\pi}{6}-i \sin \frac{\pi}{6}\right]^{100}$
$\Rightarrow(\overline{ z })^{100}=\cos \frac{50 \pi}{3}- i \sin \frac{50 \pi}{3}$
$=\cos \frac{2 \pi}{3}-i \sin \frac{2 \pi}{3}=\frac{-1-i \sqrt{3}}{2}$
$(\overline{ Z })^{100}$ lies in III ${ }^{\text {rd }}$ quadrant.

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