Complex Numbers And Quadratic Equations question 205
Question: $ \sqrt{i}= $
Options:
A) $ \frac{1\pm i}{\sqrt{2}} $
B) $ \pm \frac{1-i}{\sqrt{2}} $
C) $ \pm \frac{1+i}{\sqrt{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \sqrt{i}={{(i)}^{1/2}}={{[ \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} ]}^{1/2}} $ $ ={{[ \cos ( 2n\pi +\frac{\pi }{2} )+i\sin ( 2n\pi +\frac{\pi }{2} ) ]}^{1/2}} $ (where $ n\in I $ ) $ =[ \cos \frac{1}{2}( 2n\pi +\frac{\pi }{2} )+i\sin \frac{1}{2}( 2n\pi +\frac{\pi }{2} ) ] $ (Using De Moivre’s theorem) = $ [\cos \frac{4n\pi +\pi }{4}+i\sin \frac{4n\pi +\pi }{4}] $ Putting $ n=0 $ , 1 we get $ \cos \frac{\pi }{4}+i\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}=\frac{1+i}{\sqrt{2}} $ and $ \cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}=-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}=-( \frac{1+i}{\sqrt{2}} ) $ Therefore $ \sqrt{i}=\pm \frac{1+i}{\sqrt{2}} $ Trick: Check by squaring the options, here (c) is the square root of $ i $ because on squaring $ ( \pm \frac{1+i}{\sqrt{2}} ) $ ,we get $ i $ .
BETA
