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

I am using $z^*$ to symbolize $z$ conjugate.

How do I show no entire nonconstant function satisfies $f(z) = f(z^*)$

Thanks in advance

share|cite|improve this question

Suppose $f$ is an entire function such that $f(z)=f(z^*)$ for all $z$.

Since $f$ is holomorphic, the inverse function theorem says that $f$ is invertible near $a$ whenever $f'(a) \neq 0$. When $a$ is real, we have that $f(a+ih)=f(a-ih)$ for all real $h$ and so $f$ is not invertible near $a$. Thus $f'(a)=0$ whenever $a$ is real. Thus $f'$ is an entire function with nonisolated zeros, and hence is contantly zero. Since $f'$ is always zero, $f$ is constant.

share|cite|improve this answer
Nice argument :) – Beni Bogosel Mar 27 '12 at 15:14

fixed, but not a solution of the full question An entire function that is real on the real axis satisfies $f(z^*) = f(z)^*$. If it also satisfies $f(z)=f(z^*)$, then it satisfies $f(z) = f(z)^*$, that is, $f$ has real values everywhere. By Liouville's theorem (if the imaginary part is bounded, then the function is constant) we conclude $f$ is constant.

share|cite|improve this answer
Entire functions need not satisfy $f(z^*)=f(z)^*$. $f(z)=i$ is entire. – Chris Eagle Mar 27 '12 at 15:02
First statement is false. If $f(z)=iz$ then $f(z^*)=iz^*$ but $f(z)^*=-iz^*$. – evgeniamerkulova Mar 27 '12 at 15:04
You are right. I have to fix that... – GEdgar Mar 27 '12 at 15:07

A complex function $ƒ(x + i y) = u(x, y) + i v(x, y)$ is entire, if it is holomorphic and satisfies the Cauchy–Riemann equations: $$ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \qquad \mbox{and} \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}. \, $$ If $f(x+iy)=f(x-iy)= u(x, \pm y) + i v(x, \pm y)$, we get: $$ \frac{\partial u(x, \pm y)}{\partial x} = \frac{\partial v(x, \pm y)}{\partial y} = \pm\frac{\partial v(x, y)}{\partial y} =0, $$ therefore $f$ is constant.

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.