A) \[2,\frac{3\pi }{4}\]
B) \[4,\frac{3\pi }{4}\]
C) \[2\sqrt{2},\frac{\pi }{4}\]
D) \[2\sqrt{2},\frac{\pi }{2}\]
E) \[2\sqrt{2},\frac{3\pi }{4}\]
Correct Answer: E
Solution :
\[\frac{(1+i\sqrt{3})(2+2i)}{(\sqrt{3}-i)}=\frac{2+2i+2\sqrt{3}i-2\sqrt{3}}{(\sqrt{3}-i)}\] \[=\frac{\{(2-2\sqrt{3})+2i(1+\sqrt{3})\}}{(\sqrt{3}-i)}\times \frac{(\sqrt{3}+i)}{(\sqrt{3}+i)}\] \[=\frac{2\sqrt{3}-6+2i-2\sqrt{3}i+2\sqrt{3}i+6i-2-2\sqrt{3}}{3+1}\] \[=\frac{-8+8i}{4}=-2+2i\] \[\therefore \]Magnitude of\[\frac{(1+i\sqrt{3})(2+2i)}{(\sqrt{3}-i)}\] \[=\sqrt{4+4}=2\sqrt{2}\] and amplitude of\[\frac{(1+i\sqrt{3})(2+2i)}{(\sqrt{3}-i)}\] \[={{\tan }^{-1}}\left( \frac{2}{-2} \right)=\frac{3\pi }{4}\]
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