I was given a multi part problem to solve this. The first part is: suppose that f(z) exists and prove that there exists an entire function $g:\mathbb{C} \rightarrow \mathbb{C}$ such that $f(z) + iz = e^{ig(z)}$.

Where should I start with this?

I tried algebraically solving for $g(z)$ but ended up with a mess. I tried defining $f(z) = \sqrt{1 - z^2}$ but got stuck once I had to solve for $g(z)$ again algebraically. In a way, I need to find some explicit $g(z)$ in terms of $f(z)$ that I can use for the next part, where I'm asked to show $f(z) = \cos(g(z))$ and $z = \sin(g(z))$. I suppose this is a hint for what $g$ is, but sadly, I've never seen any problem like this before. I don't know how am I supposed to set up the appropriate equation. Any tip in the right direction is greatly appreciated.

  • $\begingroup$ if $f(z) + iz=e^{ig(z)}$ and you have to solve for $g(z)$ then just take the logarithm: $g(z) = \log(z-i\sqrt{1-z^2}).$ $\endgroup$
    – ziggurism
    Dec 17, 2023 at 14:46
  • $\begingroup$ to relate trig functions to exponential functions, remember Euler's identity that $e^{iz}=\cos z+i\sin z$ and therefore $-i\log (a+ib)=\cos^{-1}\left(\frac{a}{\sqrt{a^2+b^2}}\right)$ and similar. $\endgroup$
    – ziggurism
    Dec 17, 2023 at 14:53
  • $\begingroup$ @ziggurism: Neither $\log$ nor $\sqrt{}$ are well-defined entire functions. $\endgroup$
    – Stefan
    Dec 17, 2023 at 14:56
  • $\begingroup$ @Stefan. correct. I presume the intent of the instructions is to derive a contradiction. $\endgroup$
    – ziggurism
    Dec 18, 2023 at 0:13
  • $\begingroup$ @ziggurism: Yes, you can use that argument for the square root, but you still have to prove that $\{f(z)+iz:z\in\mathbb{C}\}$ is contained in a domain of the complex logarithm. It doesn't really matter though, the proof given in the accepted answer is so simple that it seems pointless to follow the steps of the OP's exercise. $\endgroup$
    – Stefan
    Dec 18, 2023 at 18:02

1 Answer 1


You can prove the statement directly instead.

Suppose that such an $f$ exists. Then $f(z)^2=1-z^2$. This implies $f(1)=0$. However differentiating gives $2f'(z)f(z)=-2z$. In particular for $z=1$ we get $0=-2$.


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