What I am trying to prove is that if $f$ is entire and $e^f$ is constant $f$ is constant. This is my attempt but I just can't go forward to get anything meaningful. $$e^{f(z)}=c$$ where $c\in \mathbb{C}$ is a constant. Then the set of solutions for $f(z)$ is {$\log|c|+i(Arg(c)+2k\pi)|k\in \mathbb{Z}$}. But then how do I show that $f$ is constant here after. Any help will be appreciated. Thanks
Asked
Active
Viewed 563 times
2 Answers
6
Hint: Since $c$ can't be zero, it helps to consider the derivative of both sides.
Cameron Buie
- 102,994
-
2$f'(z)e^{f(z)}=0$ since $e^f$ cannot be zero $f'=0$ which means $f$ is constant ! I hope I am correct – Heisenberg Dec 23 '14 at 12:50
-
Indeed, you are! – Cameron Buie Dec 23 '14 at 12:50
-
@Heisenberg Yes :) – Zubin Mukerjee Dec 23 '14 at 12:50
-
@CameronBuie Thanks amazes me how such simple things skip my mind. Thanks! – Heisenberg Dec 23 '14 at 12:50
4
By the chain rule, we can take the derivative of both sides to get
$$e^{f(z)} = c$$
$$f'(z)e^{f(z)}=0$$
$$f'(z)\cdot c = 0$$
$$f'(z)=0$$
This implies $f(z)$ is constant.
Zubin Mukerjee
- 17,832
-
@Heisenberg It's basically what you found based on Cameron Buie's hint, with the slight difference of substituting $c$ for $e^{f(z)}$ instead of using $e^{f(z)}\neq 0$. Same thing in the end. – Zubin Mukerjee Dec 23 '14 at 12:52
-