# Problem of the Week #88 - February 3rd, 2014

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#### Chris L T521

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No one answered this week's problem. You can find the solution below.

Proof: Let $g:\mathbb{C}^{\ast}\rightarrow\mathbb{C}$, $g(z)=f(1/z)$. Clearly, $g$ is holomorphic everywhere except the origin. We now determine what type of singularity occurs for $g$ at the origin. If $z=0$ is a removable singularity for $g$, then $g$ is bounded on a closed disk centered at $z=0$, which implies that $f$ is bounded outside this closed circle containing $z=0$. But $f$ is bounded on the closed circle, since $f$ is continuous; therefore $f$ is bounded. Therefore, since $f$ is entire and bounded, $f$ is constant by Liouville's Theorem, which contradicts the injectivity of $f$. Thus, $z=0$ is not a removable singularity for $g$.

Now suppose that $0$ is an essential singularity of $g$. Then, by the Cassorati-Weierstrass Theorem, if we chose a punctured disk at the origin $D\backslash\{0\}$, then $g(D\backslash\{0\})$ is dense in $\mathbb{C}$. This implies $f(\{|z|>r\})$ is dense in $\mathbb{C}$. But $f(\{|z|<r\})$ is open because any holomorphic mapping is open. Then $f(\{|z|>r\})\cap f(\{|z|<r\})\neq\emptyset$, which also contradicts the injectivity of $f$.

Therefore, $0$ is a pole of $g$. Since the Laurent series expansion is unique, and the principal part of $g$ is the same as the analytic part of $f$, it follows that the analytic part of $f$ has finitely many terms in it, i.e. $f$ is a polynomial. Since $f$ is injective, $f$ can have at most one root. Since $f$ can't be constant due to one of the previous cases, it follows that $f$ can only be of the form $f(z)=az+b$ where $a,b\in\mathbb{C}$ and $a\neq 0$.

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