A) 256 and \[\frac{\pi }{3}\]
B) 256 and \[\frac{2\pi }{3}\]
C) 2and \[\frac{2\pi }{3}\]
D) 256 and \[\frac{8\pi }{3}\]
Correct Answer: B
Solution :
Let \[z={{(1+i\sqrt{3})}^{8}}\] \[={{(-2)}^{6}}{{\left( \frac{1+i\sqrt{3}}{-2} \right)}^{8}}={{(-2)}^{8}}{{({{\omega }^{2}})}^{8}}\] \[={{2}^{8}}\,\,{{\omega }^{16}}={{2}^{8}}\omega \] (\[\because \] \[{{\omega }^{3}}=1\]) \[={{2}^{8}}\left( \frac{-1+i\sqrt{3}}{2} \right)\] \[={{2}^{8}}\left( \cos \frac{2\pi }{3}+i\,\,\sin \frac{2\pi }{3} \right)\] \[\therefore \] Modulus \[={{2}^{8}}=256\]and Amplitude \[=\frac{2\pi }{3}\].
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