I was given the following problem.
Consider the function $w=(z^2-4)^{1/2}$. Insert branch cuts in the z-plane such that $\pi/4>\arg{w}>-3\pi/4$. Find the equations of the branch cuts in the $z$-plane.
This is the same problem as this question, but here I completed the solution.
My solution is as follows.
We note that the principal branch($n=0$ branch) of the square root function $w=z^{1/2}$ with branch cut on the positive imaginary axis has the argument range of $-3\pi/ < \arg{w} < \pi/4$. This is because we have $-3\pi/2 < \arg{z} <\pi/2$ and therefore $-3\pi/4 <\arg{w}=\arg{z}/2 < \pi/4$.
Now, consider the intermediate function $u=z^{2}-4$. We want to find the domain of $z$ such that the range of $u$ is the entire complex plane except for the positive imaginary axis. If we could find such a domain, then by composing with the square root function, namely $w=u^{1/2}$, we have the argument range of $-3\pi/4< \arg{w} < \pi/4$.
The desired branch cut is the curve excluded from the domain.
Let $x,y$ be real numbers. Then, If $z=x+yi$, we have $u=(x+yi)^2-4=x^2-y^2-4+2xyi$. Since $u$ should not be in a positive imaginary axis, we need to exclude $z=x+yi$ such that
$\begin{cases} x^2-y^2-4=0 \\xy\geq 0\end{cases}\tag*{}$
from the domain. These are the equations of the branch cuts. Especially, the branch cut for the branch point $z=2$ is $y=\sqrt{x^2-4}(x\geq 2)$, and the branch cut for the branch point $z=-2$ is $y=-\sqrt{x^2-4}(x\leq -2)$.
The figure below shows the branch cuts on the $z$-plane.
When I got the work back, the professor wrote that he did not follow anything after "Now, consider the intermediate function...". Is there any better way to phrase this solution, or is it even correct? If it is beyond repair, what is the solution?