Showing two entire functions are equal

Asked Feb 29 '20 at 20:01

Active Feb 29 '20 at 20:01

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Let $f$ and $g$ be entire functions which satisfy $f(0)=g(0)\neq0$ and $|f(z)|\leq|g(z)|$ for all $z\in\mathbb{C}$. Show that $f=g$.

My approach: Since $f$ and $g$ are both entire, we know that $f(z)=e^{g(z)}\neq0$. Let us suppose that $f'(z)/f(z)=g(z)$. Then, $f(z)=e^{g(z)}\implies f(z)=e^{\frac{f'(z)}{f(z)}}$. Taking the log of both sides gives $log(f(z))=\frac{f'(z)}{f(z)}$ and so $f(z)\log(f(z))=f'(z)$..... and now I feel like I'm going down a rabbit hole that does not lead into Wonderland. Any thoughts?

asked Feb 29 '20 at 20:01

User7238

2,474

Also: https://math.stackexchange.com/q/613524/42969. – Martin R Feb 29 '20 at 20:06
Hi Martin! Why would the hypothesis about $f(0)=g(0)$ be important? – User7238 Feb 29 '20 at 20:07
As demonstrated in the other Q&As, $|f(z)|\leq|g(z)|$ implies $f = cg$ with some constant $c$. Then $f(0) = g(0) \ne 0$ implies that $c=1$. – Martin R Feb 29 '20 at 20:08
Ahhh.... I see.... So in my rabbit trail of nonsense, did I say anything wrong, or did I just get stuck at the last point? – User7238 Feb 29 '20 at 20:12
I do not see how you conclude that $f(z) = e^{g(z)}$. – Martin R Feb 29 '20 at 20:16
Ahlfors derives it on page 193 and 194 in Complex Analysis – User7238 Feb 29 '20 at 20:17
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If $f$ is entire without zeros then $f=e^G$ with some entire function $G$. But: 1) $f$ can have zeros, and 2) that function $G$ would be unrelated to the given function $g$. – Martin R Feb 29 '20 at 20:22
I see, thank you. Maybe it is my own stupidity, but I still don't see why $f(0)=g(0)\neq0$ would imply that $c=1$. Of course $c=1$, I just don't see why that would imply it. – User7238 Feb 29 '20 at 20:26
Substitute $z=0$ in $f(z) = cg(z)$ ... – Martin R Feb 29 '20 at 20:28
Oh, wow, yes, duh. Thank you! – User7238 Feb 29 '20 at 20:29

Showing two entire functions are equal

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