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I was workink with some peers while preparing a Complex Analysis exam and we stumbled against a problem which we cannot solve. The problem states:

Let $\Omega\subset\mathbb{C}$ an open set, and $f:\Omega\setminus\{z_0\}\to\mathbb{C}$ is holomorphic and injective. Prove that $z_0$ is either a removable singularity or a simple pole.

And it gives the hint to use Casorati-Weierstrass Theorem to show $z_0$ cannot be an essential singularity.

Here is what we managed to deduce:

First of all, WLOG $z_0=0$ and $\mathbb{D}\subset\Omega$. Let $1>\delta>0$, we have 2 possibilities.

  1. $f(D(0,\delta)\setminus\{0\})$ is bounded. In that case, $z_0$ is a removable singularity and we are done.
  2. $f(D(0,\delta)\setminus\{0\})$ is not bounded. In that case, either $z_0$ is a pole or $z_0$ is an essential singularity. If it is an essential singularity, by Casorati-Weierstrass Theorem and the Open Mapping Theorem, we know that $f(D(0,\delta)\setminus \{0\})$ is an open set dense in $\mathbb{C}$. Now let $x\in\Omega,\,x\not\in D(0,\delta)$ and any $\delta_x>0$ such that $D(x,\delta_x)\subset\Omega$ and $D(x,\delta_x)\cap (D(0,\delta)\setminus 0)=\emptyset$. Then, by the Open Mapping Theorem again, $f(D(x,\delta_x))$ is an open set containing $f(x)$. But as $f(D(0,\delta)\setminus \{0\})$ is dense in $\mathbb{C}$, this means that $f(D(x,\delta_x))\cap f(D(0,\delta)\setminus \{0\})\not=\emptyset$. But this is a contradiction with the injectivity of $f$. So $z_0$ is a pole.

However, we couldn't prove that the pole is simple. We thought about using the function $g(z)=(z-z_0)f(z)$, which might still be injective, or trying to deduce the Laurent series expansion, or even somehow the Cauchy's Formula. But we have not managed to achieve the required result.

Does anybody know how could we proceed to see that the pole is simple? Or how to give another proof that $z_0$ is either a removable singularity or a pole which by construction implies it is simple? Any help or ideas would be appreciated, thank you very much.

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1 Answer 1

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If $f$ has a multiple pole at $z_0$ then $g=1/f$ has a multiple zero at $z_0$. In particular, $g'(z_0) = 0$, which implies that $g$ (and consequently, $f$) is not injective in a neighborhood of $z_0$ (see for example Proof that 1-1 analytic functions have nonzero derivative).

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  • $\begingroup$ Thank you so much Martin, that was indeed a very simple and clean answer. $\endgroup$
    – CarlGarc27
    Jun 23, 2023 at 16:58

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