$f:\Bbb C\to\Bbb C$ is continuous, as well as analytic outside $[-1,1]$; show $f$ is entire

Question

I need to show that if $f:\Bbb C\to\Bbb C$ is continuous with $f$ analytic outside $[-1,1]$ then $f$ is an entire function. I do not know where to start.

Answer 1

Theorem: if $f(z)$ is analytic on an open rectangle $R$ and continuous on its closure, $\int_{\partial R} f(z) dz = 0$.

Note that Morera's theorem holds for $f$ on on an open connected set $U$ if the integral of $f$ around every rectangle in $U$ (with sides parallel to the $x$ and $y$ axes) is 0 (I can fill in the details if need be). By the above theorem this holds in this case: if you want to integrate $f$ around a rectangle enclosing some part of $[-1, 1]$, you can chop it into several smaller rectangles so that the portion of $[-1, 1]$ being integrated around lies on their boundaries. Thus $f$ satisfies the hypotheses of Morera's theorem and is analytic on $\Bbb{C}$.

Answer 2

It's easy using Morera's theorem (in the version that just requires the integral around triangles be $0$).