$\int_{|z|=1} \sqrt{9-z^2} \, dz$ I need to use local Cauchy theorem to find $\int_{|z|=1} \sqrt{9-z^2} \, dz$.
$\sqrt{9-z^2}$ is not analytic when $9-z^2 \in (-\infty,0]$, which means it's not analytic if $z\geq-\sqrt{3}$, is this correct? So I can't see how to enclose the path on a disk where the function is analytic to use Cauchy!
What am I doing wrong? This got me so confused.
 A: Trigger warning: I will use a phrase that is a misnomer under currently prevailing conventions.
The function $z\mapsto\sqrt{9-z^2}$ for $z\in\mathbb C$ is a "double-valued function". It might be better to write it as $z\mapsto\pm\sqrt{9-z^2}$ to emphasize that. There are two points at each of which this "function" has just one value, and those are $z=\pm3.$ Each of those is a branch point: Suppose you start at $z=3+1$ and go in a counterclockwise circle through $3+i,$ then $3+(-1),$ then $3+(-i),$ then back to $3+1,$ while the value of $\pm\sqrt{9-z^2}$ changes in a continuous way, then that value goes from $\sqrt{9-(3+1)^2} = +\sqrt 5,$ the positive square root of $5,$ to $-\sqrt 5,$ the negative square root of $5.$ If you then continue counterclockwise around the circle, then when you return the second time to $3+1,$ the value of the function returns to $+\sqrt5.$
If you go around the circle $|z|=1,$ you do not wind around either $3$ or $-3,$ so you remain on the same branch of the double-valued function $z\mapsto\pm\sqrt{9-x^2},$ and that could be either of the two branches. That function is holomorphic on an open set containing the closed disk whose boundary is that circle. Therefore the integral is $0.$
