
\begin{align*}
(a+\i b)^2 = \i & \iff a^2-b^2 + 2ab\i = \i\\
& \iff \begin{cases}
a^2-b^2=0\\
2ab = 1
\end{cases}\\
& \iff \begin{cases}
a^2 - b^2=0\\
b = \dfrac{1}{2a}
\end{cases}\\
& \iff \begin{cases}
a^2 - \left(\frac{1}{2a}\right)^2=0\\
b = \dfrac{1}{2a}
\end{cases}\\
& \iff \begin{cases}
2a^4 - 1 =0\\
b = \dfrac{1}{2a}
\end{cases}\\
& \iff \begin{cases}
a^4=\frac{1}{2}\quad\text{soit}\quad a^2=\frac{\sqrt2}{2}\quad\text{soit}\quad a=\pm\frac{\sqrt{\sqrt2}}{\sqrt2}\\
b = \dfrac{1}{2a}
\end{cases}
\end{align*}
Simplifions l'écriture de $a$ en considérant que $\sqrt2=2^{1/2}$, et donc que $\sqrt{\sqrt2}=\Big(2^{1/2}\Big)^{1/2}=2^{1/4}$.

Ainsi, $a=\pm\dfrac{\sqrt{\sqrt2}}{\sqrt2}=\pm\dfrac{2^{1/4}}{2^{1/2}}=\pm2^{-1/2}=\pm\dfrac{1}{\sqrt2}=\pm\dfrac{\sqrt2}{2}$.

Alors, $b=\pm\dfrac{1}{\sqrt2}=\pm\dfrac{\sqrt2}{2}$

\bigskip

On obtient finalement:
\[
\boxed{\sqrt\i = \pm\left(\dfrac{\sqrt2}{2}+\dfrac{\sqrt2}{2}\i\right)}
\]
