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

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

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

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

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

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

2. split $\displaystyle \oint f(z)dz = 0$ 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 $\displaystyle f$ is holomorphic on $\displaystyle \mathbb{C}\setminus\mathbb{R}$ but in order to show that $\displaystyle f$ is entire I must show it is holomorphic on $\displaystyle \mathbb{C}$ 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