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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

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Hint: Since $c$ can't be zero, it helps to consider the derivative of both sides.

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    $\begingroup$ $f'(z)e^{f(z)}=0$ since $e^f$ cannot be zero $f'=0$ which means $f$ is constant ! I hope I am correct $\endgroup$ – Heisenberg Dec 23 '14 at 12:50
  • $\begingroup$ Indeed, you are! $\endgroup$ – Cameron Buie Dec 23 '14 at 12:50
  • $\begingroup$ @Heisenberg Yes :) $\endgroup$ – Zubin Mukerjee Dec 23 '14 at 12:50
  • $\begingroup$ @CameronBuie Thanks amazes me how such simple things skip my mind. Thanks! $\endgroup$ – Heisenberg Dec 23 '14 at 12:50
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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.

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  • $\begingroup$ @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. $\endgroup$ – Zubin Mukerjee Dec 23 '14 at 12:52
  • $\begingroup$ Yep I realized thanks again $\endgroup$ – Heisenberg Dec 23 '14 at 12:53

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