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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

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

1 Answer 1

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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 $D\setminus\{0\}.$

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  • $\begingroup$ Thank you for the solution, can you please tell me if what I did is correct? $\endgroup$
    – Amit Makis
    Apr 27, 2022 at 19:41
  • $\begingroup$ In my opinion it is correct. You are applying the Morera theorem. My solution is based on MacLaurin series as $f(z)=f(0)+\sum_{n=2}^\infty a_nz^n.$ $\endgroup$ Apr 27, 2022 at 20:10

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