Proposition

Compute

$\sqrt{i}$,
$\sqrt{-i}$,
$\sqrt{1 + i}$,
$\sqrt{\frac{1 - i\sqrt{3}}{2}}$.

Solution

1

Let $a + bi = \sqrt{i}$. By squaring both sides, we obtain $a^2 = b^2$ and $2ab = 1$.

If $a = b$, then $a = \pm 1/\sqrt{2}$.
It is impossible that $a + b = 0$ because $2ab = 1 > 0$.

Therefore, $\sqrt{i} = \pm\frac{1 + i}{\sqrt{2}}$.

2

$\sqrt{-i} = \sqrt{-1}\sqrt{i} = i\sqrt{i}$. Therefore, $\sqrt{i} = \pm\frac{-1 + i}{\sqrt{2}}$.

3

By the formula given on P.4 (Ahlfors),

\[\begin{align*} \pm \Big(\sqrt{\frac{1 + \sqrt{2}{2}}{2}} + i\sqrt{\frac{-1 + \sqrt{2}}{2}}\Big). \end{align*}\]

4

Similarly,

\[\begin{align*} \pm \Big(\sqrt{\frac{1/2 + \sqrt{1/4 + 3/4}}{2}} - i\sqrt{\frac{-1/2 + \sqrt{1/4 + 3/4}}{2}}\Big) &= \pm \Big(\sqrt{3/4} - i\sqrt{1/4}\Big) \\ &= \pm \frac{\sqrt{3} - i}{2}. \end{align*}\]