how to find the branch points and cut For $f(z) = \sqrt{z^2+1}$, how can I find the branch points and cuts?
I took $z=re^{i\theta+2n\pi}$ and substitute in $f(z)$
$$\sqrt{r^2 e^{i(2\theta +4n\pi)}+e^{i 2k\pi}}=$$ 
then, I don't know how to deal with this any more
and by guessing,    
I think the branch points should be $i,-i$ and cut is $[-i,i]$
 A: Your solution is correct, but since you are guessing, I will explain it.
The values of $z$ that make the expression under the square root zero will be branch points; that is, $z = \pm i$ are branch points. Let $z - i = r_1e^{i\theta_1}$ and $z +i = r_2e^{i\theta_2}$.  Then $f(z) = \sqrt{z^2 + 1} = \sqrt{r_1r_2}e^{i(\theta_1+\theta_2)/2}$.


*

*If we don't encircle any branch point, after one revolution, $f(z)\mapsto f(z)$.

*If we encircle $z=i$ but not $z = -i$, then 
$$
\sqrt{r_1}e^{i(\theta_1+2\pi)/2} = \sqrt{r_1}e^{i\theta_1/2}e^{\pi i} = -\sqrt{r_1}e^{i\theta_1/2} 
$$
Therefore, $f(z)\mapsto -f(z)$ which is multiple valued

*Same thing happens when we encircle $z=-i$ but not $z=i$

*Lets encircle both branch points.
$$
\sqrt{r_1r_2}e^{i(\theta_1+\theta_2+2\pi+2\pi)/2} = \sqrt{r_1r_2}e^{i(\theta_1+\theta_2)/2}e^{2\pi i} = \sqrt{r_1r_2}e^{i(\theta_1+\theta_2)/2}\cdot 1
$$
So $f(z)\mapsto f(z)$ still single valued.


We could choose $[i, \infty)$ and $[-i, -\infty)$, but from item 4, we have seen traversing around both points returns the function to its original value.  Therefore, we can choose a finite branch cut, namely, $[-i, i]$.
