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?
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.