# Thread: [SOLVED] showing a complex-valued function is entire

1. ## [SOLVED] showing a complex-valued function is entire

$f$ is continuous on $\mathbb{C}$ and analytic on $\mathbb{C} \setminus \mathbb{R}$.
Prove that $f$ is an entire function.

I know that I need to apply Morera's Theorem which states that if $f$ is a continuous, complex-valued function defined on an open set $D \in \mathbb{C}$, satisfying

$\oint f(z)dz = 0$
for every closed curve $C$ in $D$, then $f$ must be holomorphic on $D$.

I guess I'm not sure how to show that $f$ satisfies $\oint f(z)dz = 0$. I thought I could use Cauchy's integral theorem to show that the function does satisfy this condition but I only know that $f$ is holomorphic on $\mathbb{C} \setminus \mathbb{R}$. Any help would be much appreciated.

2. split $\oint f(z)dz = 0$ into two integrals over two different closed contours: one that lies in the upper half plane and one in the lower. by the Cauchy theorem each of the integrals equals 0.

3. Thank you for your help. I can see how to apply Cauchy's integral theorem in this manner, but in conjunction with Morera's theorem it seems as though I am only showing that $f$ is holomorphic on $\mathbb{C}\setminus\mathbb{R}$ but in order to show that $f$ is entire I must show it is holomorphic on $\mathbb{C}$ Is there something I am missing here?

4. take *any* contout in the complex plane and divide it into two by splitting along the real line. you show that the integral is zero by Cauchy, and that implies f is entire