If z∈C then what is z2?

Step 1
I'm not sure why specifying ${\displaystyle z\in \mathbb{C}}$ is significant.
If ${\displaystyle z=a+bi}$
then
${\displaystyle z^2=a^2+2abi-b^2}$
Any value (in ${\displaystyle \mathbb{C},)\hat{z}}$ for which
${\displaystyle {\hat{z}}^2=a^2+2bi-b^2}$
should be a square root of z
The two possible $\hat{z}$ values are
${\displaystyle \hat{z}=a+bi=z}$
and
${\displaystyle \hat{z}=-a-bi=-z}$

Step 1
I would take the principal square root of a complex number z, to be:
${\displaystyle \sqrt{z}=\sqrt{\left|z\right|}\left(cis\left(\frac{1}{2}arg\left(z\right\}\right)\right)}$
where ${\displaystyle arg\left(z\right)}$ is the principal argument of z, which some take to be in ${\displaystyle [0,2\pi )}$ and others take to be in ${\displaystyle (\pi ,\pi ]}$
So for radicand ${\displaystyle z^2}$, I would take
${\displaystyle \sqrt{z^2}=\sqrt{\left|z^2\right|}\left(cis\left(\frac{1}{2}arg\left(z^2\right)\right)\right)}$
where $\arg (z^2)$ is the principal argument of ${\displaystyle z^2}$, which some take to be in ${\displaystyle [0,2\pi )}$ and others take to be in ${\displaystyle (-\pi ,\pi ]}$
Given a choice, would prefer ${\displaystyle arg}$ in ${\displaystyle (-\pi ,\pi ]}$