This is a problem from Complex Variable (Conway's book) 2nd ed.
(Section 4.4) 9. Show that if $f: \mathbb{C}\to\mathbb{C}$ is a continuous function such that $f$ is analytic off $[-1,1]$ then $f$ is an entire function.
I already have a solution by Morera's theorem that split this problem in 5 cases. I think this solution is too long and I'm trying to solve this using a different approach. Any ideas ?
Best Answer
Like Daniel Fischer, I used a Cauchy integral formula approach. Let $\gamma (t) = 2e^{it}, 0\le t \le 2\pi.$ For small $\epsilon>0$ let $R_\epsilon=[-1-\epsilon,1+\epsilon]\times [-\epsilon,\epsilon].$ Consider $\partial R_\epsilon$ as a contour in the natural way, oriented clockwise.
Let $z \in D(0,2)\setminus [-1,1].$ Then $z\notin R_\epsilon$ for small $\epsilon >0.$ For such $\epsilon$ we have by Cauchy
$$f(z) = \frac{1}{2\pi i}\left ( \int_{\gamma + R_\epsilon} \frac{f(w)}{w-z}\,dw\right ).$$
Verify that the integral over $R_\epsilon \to 0$ as $\epsilon\to 0$ (here using the continuity of $f$). Thus we have
$$f(z) = \frac{1}{2\pi i}\left ( \int_\gamma \frac{f(w)}{w-z}\,dw\right )$$
for all $z \in D(0,2)\setminus [-1,1].$ But the last integral defines a holomorphic function in all of $D(0,2).$ Simply by continuity, $f$ equals this integral in all of $D(0,2)$ and we're done.