Even holomorphic function on the punctured disk has a primitive

Question

Let $f$ such that $f$ is holomorphic on $\{z|0<|z|<1\}$, and $f$ is even.

I need to show that f has a primitive. Any ideas?

Answer 1

Write $f$ in the form of Laurent series around $0$. You get the form $f(z)=\sum_{n=-\infty}^\infty a_nz^n$. Now, if we prove that $a_{-1}=0$ then $F(z)=\sum_{n=-\infty}^{\infty}\frac{a_n}{n+1}z^{n+1}$ is a primitive function of $f$. So we just have to prove that $a_{-1}=0$ and we are done. And here we are going to use that $f$ is even. We know that $\sum_{n=-\infty}^\infty a_nz^n=f(z)=f(-z)=\sum_{n=-\infty}^\infty a_n(-z)^n$. So if we define $g(z)=f(z)-f(-z)$ then $g$ is the zero function and its Laurent expansion is $\sum_{n=-\infty}^\infty a_{2n+1}z^{2n+1}$. Because Laurent expansion is unique we conclude that $a_{2n+1}=0$ for all $n\in\mathbb{Z}$. So $a_{-1}=0$.

Answer 2

Let $F$ be a primitive of $f\restriction_{\Bbb D\setminus(-\infty,0]}$. Then $z\mapsto F(z)+F(-z)$ is constant where defined, i..e, constant on the upper half disk and constant on the lower half disk. Clearly, the constants are negatives of each other. By adding a suitable constant to $F$, we can achieve that $F(z)+F(-z)=0$ thoughout, i.e., $F$ is odd. If we defien $F(z)=-F(-z)$ on $(-1,0)$, we obtain a continuos (and necesarily holomorphic) function on the whole punctured disk.

Answer 3

Let $0 < r < 1$. Then we have $$\int_{\partial B_r(0)} f(z)\, dz = \int_0^{2\pi} f(re^(it)ire^{it}\, dt = \int_0^{\pi} f(re^{it})ire^{it}\, dt + \int_\pi^{2\pi} f(re^{it})ire^{it}\, dt.$$ For the second integal we have $$ \int_\pi^{2\pi} f(re^{it})ire^{it}\, dt = \int_0^{\pi} f(re^{i(t+ \pi)})ire^{i(t+ \pi) }\, dt = -\int_0^{\pi} f(-re^{it})ire^{it }\, dt= -\int_0^{\pi} f(re^{it})ire^{it }\, dt,$$ by the evenness of $f$. By homotopy, every simple loop avoiding 0 integrates to 0. Now pick a any point $z_0$ in the punctured disk. Define $F(z) = \int_\gamma f(z)\, dz,$ where $\gamma$ is a path connecting $z_0$ to $z$. This is well-defined because of the fact for that every loop in the punctured disk, the integral along the loop integrates to zero.