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
Correct Answer: C
Solution :
\[\sqrt{i}={{(i)}^{1/2}}={{\left[ \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right]}^{1/2}}\] \[={{\left[ \cos \left( 2n\pi +\frac{\pi }{2} \right)+i\sin \left( 2n\pi +\frac{\pi }{2} \right) \right]}^{1/2}}\] (where \[n\in I\]) \[=\left[ \cos \frac{1}{2}\left( 2n\pi +\frac{\pi }{2} \right)+i\sin \frac{1}{2}\left( 2n\pi +\frac{\pi }{2} \right) \right]\] (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 \[\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}=-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}=-\left( \frac{1+i}{\sqrt{2}} \right)\] 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\[\left( \pm \frac{1+i}{\sqrt{2}} \right)\],we get \[i\].
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