Does there exist a branch of $(a^2 - z^2)^{1/2}$ holomorphic on $\mathbb{C}\setminus [-a,a]$?

Question

To compute the integral

\begin{align*} \int_{-a}^a (a^2-x^2)^{1/2} \, dx \end{align*}

(and practice contour integration) I am trying to define a branch of the integrand with branch cut $[-a,a]$, and then eventually find the integral around some circular contour containing the interval of integration and then "collapsing" the contour down to the interval. But I am stuck on how to define this branch.

My trouble is that if I start by defining a branch of $z\mapsto z^{1/2}$, then the function we get for $(a^2-z^2)^{1/ 2}$ won't have the desired cut. Is there another way to define a branch of $(a^2-z^2)^{1/2}$?

Answer 1

Yes, there exists such a branch. The most natural way to define it is to write it as $$f(z) = iz\sqrt{1-\frac{a^2}{z^2}}$$ The quantity $\frac{a^2}{z^2}$ is a real number in the interval $[1,\infty)$ if and only if $z$ is in the segment $[-a,a]$. This definition, using the usual branch of the square root, therefore works everywhere outside that cut.

Answer 2

You don't need to use contour integration to do this integral. The integral is of the function $x^2+y^2=a^2$ (a circle centre $(0,0)$ radius $a$). So the area is just given by $$\frac12\pi a^2$$