Complex Numbers And Quadratic Equations question 828
Question: $ \frac{1+7i}{{{(2-i)}^{2}}}= $ [Roorkee 1981]
Options:
A) $ \sqrt{2}( \cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4} ) $
B) $ \sqrt{2}( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} ) $
C) $ ( \cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4} ) $
D) None of these
Show Answer
Answer:
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}[ \cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4} ] $ Aliter: $ | \frac{1+7i}{{{(2-i)}^{2}}} |=| \frac{1+7i}{3-4i} |=\sqrt{2} $ and $ arg( \frac{1+7i}{3-4i} )={{\tan }^{-1}}7-{{\tan }^{-1}}( -\frac{4}{3} ) $ $ ={{\tan }^{-1}}7+{{\tan }^{-1}}\frac{4}{3}=\frac{3\pi }{4} $
$ \therefore \frac{1+7i}{{{(2-i)}^{2}}} $ $ =\sqrt{2}( \cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4} ) $
BETA
