A) I quadrant
B) II quadrant
C) III quadrant
D) IV quadrant
Correct Answer: C
Solution :
\[z=\frac{1+i\sqrt{3}}{\sqrt{3}+i}\]\[\Rightarrow z=\frac{1+i\sqrt{3}}{\sqrt{3}+i}\times \frac{\sqrt{3}-i}{\sqrt{3}-i}\] \[\Rightarrow \] \[z=\frac{\sqrt{3}+3i-i+\sqrt{3}}{3+1}\]\[=\frac{2(\sqrt{3}+i)}{4}\] \[\Rightarrow \] \[z=\frac{\sqrt{3}+i}{2}=\left[ \cos \frac{\pi }{6}+i\sin \frac{\pi }{6} \right]\] Now \[\bar{z}=\cos \frac{\pi }{6}-i\sin \frac{\pi }{6}\] Þ \[{{(\bar{z})}^{100}}={{\left[ \cos \frac{\pi }{6}-i\sin \frac{\pi }{6} \right]}^{100}}\] Þ \[{{(\bar{z})}^{100}}=\cos \frac{50\,\pi }{3}-i\sin \frac{50\,\pi }{3}\]\[=\,\cos \frac{2\pi }{3}-i\sin \frac{2\pi }{3}\] \[{{(\bar{z})}^{100}}\] lies in III quadrant.
You need to login to perform this action.
You will be redirected in
3 sec
