Here it is another solution for the sake of curiosity.

Let $z = x + yi$. Then we have the following equation:

\begin{align*} z^{2} + i = 0 & \Longleftrightarrow (x+yi)^{2} = x^{2} - y^{2} + 2xyi = -i\\\\ & \Longleftrightarrow \begin{cases} x^{2} - y^{2} = 0\\\\ 2xy = -1 \end{cases}\\\\ & \Longleftrightarrow \begin{cases} x = -y\\\\ 2y^{2} = 1\\ \end{cases}\\\\ & \Longleftrightarrow \begin{cases} x = -\dfrac{1}{\sqrt{2}}\\\\ y = +\dfrac{1}{\sqrt{2}} \end{cases}; \begin{cases} x = +\dfrac{1}{\sqrt{2}}\\\\ y = -\dfrac{1}{\sqrt{2}} \end{cases} \end{align*}

Hopefully this helps!

Here it is another solution for the sake of curiosity.

Let $z = x + yi$, where $x,y \in \mathbb{R}$. Then we have the following equation:

\begin{align*} z^{2} + i = 0 & \Longleftrightarrow (x+yi)^{2} = x^{2} - y^{2} + 2xyi = -i\\\\ & \Longleftrightarrow \begin{cases} x^{2} - y^{2} = 0\\\\ 2xy = -1 \end{cases}\\\\ & \Longleftrightarrow \begin{cases} x = -y\\\\ 2y^{2} = 1\\ \end{cases}\\\\ & \Longleftrightarrow \begin{cases} x = -\dfrac{1}{\sqrt{2}}\\\\ y = +\dfrac{1}{\sqrt{2}} \end{cases}; \begin{cases} x = +\dfrac{1}{\sqrt{2}}\\\\ y = -\dfrac{1}{\sqrt{2}} \end{cases} \end{align*}

Hopefully this helps!