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Let $f :\mathbb{C} \longrightarrow \mathbb{C}$ be an analytic injective function with $f(0)=0$, let $r>0$. Since $\partial B_r(0)$ is a compact set, then $|f(z)|$ assumes a minimum on $\partial B_r(0)$, say $\varepsilon:=\underset{z \in \partial B_r(0)}{\min} |f(z)|>0$ by injectivity.

Assume that there is $z_0$ with $|z_0|>r$ such that $|f(z_0)|< \varepsilon$. Does then follow with the maximum principle that $f$ is constant? My idea is that the function $z \mapsto |f(z)|$ would then have a maximum inside the ring $\{z \in \mathbb{C}: r < |z| < |z_0| \}$ because of its continuity.

How do I make this argument more rigorous or contradict if it is false?

The Lion King
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There is a flaw in your argument because $|f(z_0)|< \varepsilon$ with $|z_0|>r$ does not imply that the maximum of $|f(z)|$ on some larger circle is $< \varepsilon$.

As a simple example we can choose $f(z) = z-2$ and $r=1$. Then $$ \varepsilon =\min_{z \in \partial B_r(0)} |f(z)| = 1 $$ and $|f(2)| = 0 < \epsilon$.

As a side note, the only entire and injective functions are linear functions of the form $f(z) = az+b$, see for example Proving that all entire & injective functions take the form $f = ax + b$? or $f$ be an entire function, that is also injective. Then $f$ take the form $f(z) = az+b$ with $a, b \in \Bbb C$ and $a \neq 0$.

Martin R
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