Question

What is the square root of $i$, where $i=\sqrt{-1}$?

• A. $\frac{1+i}{2}$
• B. $\frac{1-i}{2}$
• C. $\frac{1+i}{\sqrt{2}}$
• D. None of the above

Answer 1

The correct option is C $\frac{1+i}{\sqrt{2}}$

Let the square root of $i$ be $a+ib$

$⟹{(a+ib)}^2=i$

$⟹a^2+2abi-b^2=i$

$⟹a^2-b^2=0,2ab=1$

Now, $a^2-b^2=0$

$⟹(a+b)(a-b)=0$

$⟹a\ne -b,a=b$

$2ab=1$

$⟹2a^2=1$

$⟹a=\pm \frac{1}{\sqrt{2}},b=\pm \frac{1}{\sqrt{2}}$

$\therefore \sqrt{i}=a+ib=\frac{1+i}{\sqrt{2}},\frac{1-i}{\sqrt{2}}$