# Proving an analytic function in a domain that is not simply connected has an antiderivative

Let $$f$$ be analytic in $$\mathbb{D}$$ s.t $$f'(0)=0$$. Prove that $$\frac{f(z)}{z^2}$$ has an antiderivative on $$\mathbb{D}\setminus\{0\}$$.

My attempt: Let $$\gamma$$ be a closed curve in $$\mathbb{D}$$. If $$0\not\in\ int(\gamma)$$ then by Cauchy-Goursat, $$\int_{\gamma}\frac{f(z)}{z^2}dz=0$$. Else, if $$0\in int(\gamma)$$, then by Cauchy's integral formula: $$\int_{\gamma}\frac{f(z)}{z^2}dz=2\pi if'(0)=0$$ Either way, the integral of $$\frac{f(z)}{z^2}$$ on every closed curve is $$0$$ and $$f$$ is continuous on $$\mathbb{D}\setminus\{0\}$$ and hence has antiderivative there.

Is this correct or is my use of the theorem wrong? Any help would be appreciated

• Does $D$ denote the open unit disc ? Apr 27, 2022 at 19:22
• @RyszardSzwarc Yes, in my notation this is the unit disc Apr 27, 2022 at 19:25

The function $$g(z):={f(z)-f(0)\over z^2}$$ extends to a holomorphic function on $$D.$$ We have $${f(z)\over z^2}={f(0)\over z^2} + g(z)$$ Let $$G(z)$$ denote an antiderivative of $$g(z).$$ Then the function $$-{f(0)\over z} + G(z)$$ is an antiderivative of $$f(z)/z^2$$ on $$D\setminus\{0\}.$$
• In my opinion it is correct. You are applying the Morera theorem. My solution is based on MacLaurin series as $f(z)=f(0)+\sum_{n=2}^\infty a_nz^n.$ Apr 27, 2022 at 20:10