For $z_0\in\mathbb{C},\rho>0$ let $B(z_0,\rho)=\{z\in\mathbb{C}:|z-z_0|<\rho\}$.

Suppose that $f$ is holomorphic in $B(0,1)\setminus\{0\}$, $f$ is unbounded in $B(0,1/2)$ and $f$ is injective.

Prove that $z=0$ is a simple pole of $f$.

First of all, $z=0$ can not be a removable singularity since then we can remove it to obtain a function $\tilde{f}$ that is holomorphic in $B(0,1)$, hence bounded, hence the original $f$ is bounded, contradiction.

Now, from Picard's great theorem we get that if $z=0$ is an essential singularity, then every complex number (with at most a single exception) is assumed by $f$, near $z=0$, infinitely often. Contradicting the fact that $f$ is injective.

So, it is true that $z=0$ is a pole of $f$. If $z=0$ is a pole of order $m>1$ of $f$, then $z=0$ is a zero of multiplicity $m$ of $\frac{1}{f}$ and so, by the local mapping theorem, $\frac{1}{f}$ is not injective, so $f$ is not injective. Therefore, $z=0$ must be a simple pole.

My problem is the use of Picard's great theorem, can we use a weaker argument to eliminate the second case?

  • $\begingroup$ Unlikely that Picard's can be avoided. $\endgroup$ – Kavi Rama Murthy Jun 18 '18 at 9:31

A Casorati-Weierstrass type argument is sufficient: $$ G = f \left(\{ z \mid \frac 12 < |z| < 1 \} \right) $$ is open and therefore contains an open disk $B(w_0, \varepsilon)$.

Since $f$ is injective, $f(z) \notin B(w_0, \varepsilon)$ for $|z| < \frac 12$, which means that $$ \frac{1}{f(z) - w_0} $$ is bounded near $z=0$, and therefore has a removable singularity.


You can prove it using the Casoratti-Weierstrass theorem. If $f$ has an essential singularity at $0$ and if $V$ is a neighbourhood of $0$, then $f(V\setminus\{0\})$ is dense. Therefore, $f|_{V\setminus\{0\}}$ cannot be injective.


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