# No Entire Nonconstant function satisfies $f(z) = f(z^*)$

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

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

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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.

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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.

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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.

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