A) \[\sqrt{2}\left( \cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4} \right)\]
B) \[\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)\]
C) \[\left( \cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4} \right)\]
D) None of these
Correct Answer: A
Solution :
\[\frac{1+7i}{{{(2-i)}^{2}}}=\frac{(1+7i)}{(3-4i)}\frac{(3+4i)}{(3+4i)}=\frac{-25+25i}{25}=-1+i\] Let \[z=x+iy=-1+i\] \[\therefore r\cos \theta =-1\]and \[r\sin \theta \]=1 \[\therefore \theta =\frac{3\pi }{4}\]and \[r=\sqrt{2}\] Thus \[\frac{1+7i}{{{(2-i)}^{2}}}=\sqrt{2}\left[ \cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4} \right]\] Aliter: \[\left| \frac{1+7i}{{{(2-i)}^{2}}} \right|=\left| \frac{1+7i}{3-4i} \right|=\sqrt{2}\] and \[arg\left( \frac{1+7i}{3-4i} \right)={{\tan }^{-1}}7-{{\tan }^{-1}}\left( -\frac{4}{3} \right)\] \[={{\tan }^{-1}}7+{{\tan }^{-1}}\frac{4}{3}=\frac{3\pi }{4}\] \[\therefore \frac{1+7i}{{{(2-i)}^{2}}}\]\[=\sqrt{2}\left( \cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4} \right)\]
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