Branch cut for $\sqrt{z^2-a^2}$ in the lower half of the complex plane (instead of $[-a,a]$) - How does this change the function?

Question

I have given a function $\sqrt{z^2-a^2}$ with $a>0$.

At first i have chosen the branch cut on the real axis at $-a<z<a$: $$f(z)=\sqrt{z-a}\sqrt{z+a}=\sqrt{|z-a||z+a|} \exp({i\frac{\theta_1+\theta_2}{2}})$$

My question is: What happens, if i turn the branch cuts by $\frac{\pi}{2}$ to $\{ z=\pm a -iy,\ y>0\}$? And how does this change the (value of the) function?

Change of the branch cut(s)

Naively i would assume for the square root function $\sqrt{z} \rightarrow e^{i \pi /4}\sqrt{z}=\sqrt{iz}$ , which leads to $$f(z) \rightarrow g(z)=i\sqrt{z-a} \sqrt{z+a}=\sqrt{{a^2-z^2}}$$

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This problem is part of an exercise: By changing the branch cuts $$R(z)=A\left[z-\sqrt{z^2-a^2}\right] \rightarrow \tilde{R}(z)$$ $$h(z)=\frac{1}{z-\epsilon-\tilde{R}(z)}$$ and the function $h(z)$ has a pole at $z=E_0-i \Gamma /2$.