Proving a certain complex valued function is constant.

Suppose you have a differentiable $f:\mathbb{C}\to\mathbb{C}$ such that $f(z)\neq 0$ for any $z$.

My question is, if $\lim_{z\to\infty}f(z)$ exists and is nonzero, why does that imply that $f$ is actually constant? Thanks for any explanation.

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en.wikipedia.org/wiki/… – Graphth Mar 28 '12 at 1:56

Because $f$ is nonzero and $\lim_{z\to\infty}f(z)$, you get that $\frac{1}{f(z)}$ is entire and bounded.

The reason why $\frac{1}{f(z)}$ is bounded is simple: $\lim_{z \to \infty} \frac{1}{f(z)}$ exists and is finite. Hence, there exists an $M$ and some $R>0$ so that

$$|\frac{1}{f(z)}| < M \forall |z|>R \,.$$

By continuity $\frac{1}{f(z)}$ is also bounded on the compact set $|z| \leq R$.

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Since you have $f$ differentiable then his real part and imaginary part are Harmonic ($\Delta f=0$ or $f_{xx}+f_{yy}=0$, and all Harmonic defined on $\mathbb{R}$ bounded functions are constants!

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You have an entire function. Liouville's theorem says that bounded entire functions are constant. Maximum modulus theorem says that the maximum cannot occur properly in the domain, so must occur at $\infty$. And same for the minimum. So $f$ is bounded.

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