A) 256 and\[\frac{\pi }{3}\]
B) 256 and\[\frac{2\pi }{3}\]
C) 2 and\[\frac{2\pi }{3}\]
D) 256 and\[\frac{8\pi }{3}\]
Correct Answer: B
Solution :
Let \[z={{(1+i\sqrt{3})}^{8}}\] \[={{(-2)}^{8}}{{\left( \frac{1+i\sqrt{3}}{-2} \right)}^{8}}={{(-2)}^{8}}{{({{\omega }^{2}})}^{8}}\] \[={{2}^{8}}{{\omega }^{16}}={{2}^{8}}\omega \] \[(\because {{\omega }^{3}}=1)\] \[={{2}^{8}}\left( \frac{-1+i\sqrt{3}}{2} \right)\] \[={{2}^{8}}\left( \cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3} \right)\] \[\therefore \]Modulus\[={{2}^{8}}=256\] and Amplitude\[=\frac{2\pi }{3}\].
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