Complex Numbers And Quadratic Equations question 254
Question: If $ z=\frac{\sqrt{3}+i}{2} $ , then the value of $ z^{69} $ is [RPET 2002]
Options:
A) $ -i $
B) $ i $
C) 1
D) $ -1 $
Show Answer
Answer:
Correct Answer: A
Solution:
Given that $ z=\frac{\sqrt{3}+i}{2}=\frac{\sqrt{3}}{2}+\frac{1}{2}i $
$ \Rightarrow iz=-\frac{1}{2}+i\frac{\sqrt{3}}{2}=\omega $ Now $ z^{69}={z^{4(17)}}z={{(iz)}^{4(17)}}z={{(\omega )}^{68}}z,(\because i^{4n}=1) $ $ =\frac{{{\omega }^{69}}}{i}=\frac{{{({{\omega }^{3}})}^{23}}}{i}=\frac{1}{i}=-i $ Aliter: $ z=\frac{\sqrt{3}}{2}+i\frac{1}{2}=\cos \frac{\pi }{6}+i\sin \frac{\pi }{6} $
Þ $ z^{69}={{( \cos \frac{\pi }{6}+i\sin \frac{\pi }{6} )}^{69}}=\cos \frac{69\pi }{6}+i\sin \frac{69\pi }{6} $ $ =\cos ( 11\pi +\frac{\pi }{2} )+i\sin ( 11\pi +\frac{\pi }{2} )=0+i(-1)=-i $ .
BETA
