If a = sqrt{2i},then which of the following i | vedclass

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(A) Given $a = \sqrt{2i}$.We can write $2i$ in polar form as $2(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})$.Then $a = \sqrt{2} (\cos \frac{\pi}{2} + i

Vedclass · Accepted Answer

$a = 1 + i$

Vedclass · Answer

(A) Given $a = \sqrt{2i}$.We can write $2i$ in polar form as $2(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})$.Then $a = \sqrt{2} (\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})^{1/2}$.Using De Moivre's theorem,$a = \sqrt{2} (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$.$a = \sqrt{2} (\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}) = 1 + i$.Alternatively,squaring the options: $(1+i)^2 = 1^2 + i^2 + 2i = 1 - 1 + 2i = 2i$. Thus,$a = 1+i$ is correct.

If $a = \sqrt{2i}$,then which of the following is correct?

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