What is the square root of i?

Solution:

Let \(z = a + bi\) be the complex number which is a square root of i, that is

\(z^{2} = (a + bi)^{2} = a^{2} – b^{2} + 2abi = i\)

Equating real and imaginary parts, we have

\(\begin{cases} a^{2} – b^{2} & = 0,\\ 2ab & = 1.\end{cases}\)

The two real solution to this pair of equations are

\(\begin{cases} a = \frac{1}{\sqrt{2}},\\ b = \frac{1}{\sqrt{2}},\end{cases}\)

and

\(\begin{cases} a = -\frac{1}{\sqrt{2}},\\ b = -\frac{1}{\sqrt{2}}.\end{cases}\)

The two square root of i therefor are \(\pm\frac{1}{\sqrt{2}}(i + 1)\).