A) \[a=1+i\]
B) \[a=1-i\]
C) \[a=-(\sqrt{2})i\]
D) None of these
Correct Answer: A
Solution :
We have \[a=\sqrt{2i}=\sqrt{2\,}{{i}^{1/2}}=\sqrt{2}{{\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)}^{1/2}}\] \[=\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)=\sqrt{2}\left( \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i \right)=1+i\] Trick: Check with options.
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