0

Let $f(z)= u+ iv$ where $u$, $v$ are real and imaginary parts of $f$ respectively and $f$ is entire function. If $\{u(x,y)+v(x,y): z=x+iy\in \mathbb{C} \}$is bounded then function is constant.

Here is what I tried. Consider$g(z)=f(z)-if(z)=(u+v)-i(u-v)$. Now linear combination of two entire function is entire. So $g(z)$ is entire and real part of $g$ is bounded. So $g$ is constant and hence $f$ is also constant. Am I correct? Thank you

ramanujan
  • 1,584

1 Answers1

2

Another approach:

We have the identity $$\left\vert e^z \right \vert=e^{\Re z}$$

Let $g(z)=(1-i)f(z)=(1-i)(u(x,y)+iv(x,y))$.

We have $$|\exp g(z)|=\left\vert e^{(1-i)f(z)}\right\vert=e^{u+v}\le e^M$$

Since $e^{g(z)}$ is entire and bounded, by Liouiville’s theorem $$e^{g(z)}=C\implies g(z)=\text{constant}\implies f(z)=\text{constant}$$

which completes the proof.

Szeto
  • 11,159