Suppose that $z = x + iy$, with $y > 0$. Show that there are positive real numbers $u$ and $v$ with $2u^{2} = |z| + x$ and $2v^{2} = |z| - x$ Suppose that $z = x + iy$, with $y > 0$. Show that there are positive real numbers $u$ and $v$ with $2u^{2} = |z| + x$ and $2v^{2} = |z| - x$. Calculate $(u+iv)^{2}$. Show that $z$ has exactly two complex square roots. Show that the same holds when $y < 0$.
MY ATTEMPT
Let us solve this problem backwards:
\begin{align*}
\begin{cases}
2u^{2} = |z| + x\\\\
2v^{2} = |z| - x
\end{cases} \Longleftrightarrow
\begin{cases}
|z| = u^{2} + v^{2}\\\\
x = u^{2} - v^{2}
\end{cases}
\end{align*}
Consequently, one has that
\begin{align*}
|z|^{2} = (u^{2} + v^{2})^{2} = x^{2} + y^{2} = (u^{2} - v^{2})^{2} + y^{2} & \Longleftrightarrow u^{4} + 2u^{2}v^{2} + v^{4} = u^{4} - 2u^{2}v^{2} + v^{4} + y^{2}\\\\
& \Longleftrightarrow y^{2} = 4u^{2}v^{2}
\end{align*}
Thus it suffices to choose $x = u^{2} - v^{2}$ and $y = 2uv$, where $u,v\in\textbf{R}_{\geq0}$.
Thus we conclude that
\begin{align*}
(u+iv)^{2} = u^{2} - v^{2} + 2uvi = x + yi = z
\end{align*}
If $y < 0$, it suffices to take $y = -2uv$.
Finally, I did not get what it means to show that $z$ has exactly two complex roots. Any help on this?
If there is a nice way to approach it, I'd be grateful to know. Any contribution is appreciated.

I think you accidentally solved for $x$ and $y$ instead of $u$ and $v$.  Remember that $z = x+iy$ is fixed, so we're looking for $u$ and $v$ in terms of $x$ and $y$.  I think you can probably reuse most of the work you already have for that, though.
All they mean by showing $z$ has two complex roots is that since you showed that $(u+iv)^2 = z$, we can also write $\sqrt{z} = u+iv$.  For the second complex root, you can just take use that $(-(u+iv))^2 = (u+iv)^2 = z$.
