A) \[-i\]
B) \[i\]
C) 1
D) \[-1\]
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 \[{{z}^{69}}={{\left( \cos \frac{\pi }{6}+i\sin \frac{\pi }{6} \right)}^{69}}=\cos \frac{69\pi }{6}+i\sin \frac{69\pi }{6}\] \[=\cos \left( 11\pi +\frac{\pi }{2} \right)+i\sin \left( 11\pi +\frac{\pi }{2} \right)=0+i(-1)=-i\].
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