I've been given a question regarding branch cuts, and specifically to find them I've followed how to find the branch points and cut
and this suggests for what ever points of our function f(z) that f(z*) = 0, in that example, $\pm i$ we can consider the three intervals $[-\infty,-i], [-i,i]~\&~[i, \infty]$ in their method they define values of $z$ accordingly and encircle each branch point, but by doing so, they find either the sign of $f(z)$ flips or stays the same.
but surely in their defining of $z-i$ and $z+i$ their encircling of $2\pi$ will always yield the same results,
for example, if the branch points were $1$ and $-2$, we would define $z-1 = r_1 e^{i\theta_1}$ and $z+2 = r_2 e^{i\theta_2}$
then encircling $z-1$ gives $$\sqrt{r_1} e^{i(\theta_1+2 \pi)/2} = \sqrt{r_1} e^{i\theta_1}e^{i\pi} = -\sqrt{r_1} e^{i\theta_1}$$
which by the above logic then we'd choose $[-2,1]$ to be the branch
but by Complex function, branch-cut method they essentially suggest to find the points in which for $f(x) =e^{\log{g(x)}}$ ,$g(x*) = 0$ in order to show points of discontinuity if the log function was real.
so in the above question we would have
$$\sqrt{(z+2)(z-1)} = e^{\log{\sqrt{(z+2)(z-1)}}} = e^{\frac{1}{2} \log{(z+2)(z-1)}} $$
which is negative between [-2,1]
either way that gives us the three regions and then we cut away as required?
as you can see i'm kind of confused about branch cuts so any help would be greatly appreciated.
Thanks for taking the time to read this.
