A) 2
B) 4
C) 8
D) 16
E) 32
Correct Answer: D
Solution :
\[\because \]\[(\sqrt{3}-i)=2\left( \cos \frac{\pi }{6}-i\sin \frac{\pi }{6} \right)\] \[\therefore \]\[{{(\sqrt{3}-i)}^{50}}={{2}^{50}}{{\left( \cos \frac{\pi }{6}-i\sin \frac{\pi }{6} \right)}^{50}}\] \[={{2}^{50}}\left( \cos \frac{\pi }{3}-i\sin \frac{\pi }{3} \right)\] \[={{2}^{50}}\left( \frac{1}{2}-i\frac{\sqrt{3}}{2} \right)\] But\[{{(\sqrt{3}-i)}^{50}}={{2}^{48(x+iy)}}\] \[x=2\]and \[y=2\sqrt{3}\] \[\therefore \] \[{{x}^{2}}+{{y}^{2}}=4+12=16\]
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