Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
2… – Graphth Mar 28 '12 at 1:56
up vote 2 down vote accepted

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 $|z| \leq R$.

share|cite|improve this answer

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!

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.