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Question: Suppose that functions $f$ and $g$ are analytic on the disc $A = \{z : |z| < 2\}$ and that neither $f(z)$ nor $g(z)$ is ever $0$ for $z \in A$.

$\frac{f'(\frac{1}{n})}{f(\frac{1}{n})} = \frac{g'(\frac{1}{n})}{g(\frac{1}{n})}$ for $n=1,2, ... ,$

show that there is a constant $c$ such that $f(z) = cg(z)$ for all $z \in A$.

Comments: This is one of the more theoretical problems from an old complex analysis exam. I am not sure which theorems I am supposed to use here. Am I supposed to take the integrals of both sides over the same curve and use the root-pole counting theorem? All input very much appreciated, so far everyone here has been of great help.

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2 Answers 2

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Since $f$ and $g$ are holomorphic and never zero, $\frac{f'}{f}$ and $\frac{g'}{g}$ are again holomorphic. We know that $\frac{f'}{f} = (\log(f))'$ and so what we really have is that $(\log(f))' = (\log(g))'$ on the set $\{\frac{1}{n}: n\in\mathbb{N}\}$. There is a theorem that says that if two holomorphic functions agree on a set of points with an accumulation point, then they are equal. Therefore we have that $\log(f) = \log(g) + b$ for some $b\in\mathbb{C}$. Can you take it from here?

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  • $\begingroup$ If there are any issues with my LaTeX, please correct it. MathJax is not rendering the LaTeX code for me at all so I can't see if there are any problems. $\endgroup$ Aug 29, 2013 at 1:20
  • $\begingroup$ Your LaTeX is fine and so is your solution, thank you. We'll just set exp(b)=c and then we're done. Do you know what this theorem you used is called? I'm curious if I can find it in my textbook along with a proof (Basic Complex Analysis). $\endgroup$
    – Sid
    Aug 29, 2013 at 1:49
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    $\begingroup$ @Sid The Identity Theorem. It's proved in Ahlfors' book (and many others). $\endgroup$
    – Potato
    Aug 29, 2013 at 1:52
  • $\begingroup$ @Potato Thanks! I couldn't remember it. $\endgroup$ Aug 29, 2013 at 1:54
  • $\begingroup$ I do not like this answer as you do not deal with the branch cut issue of the log. It cannot be defined per say on the branch cut but this is true for all elements of $z$ in $A$. $\endgroup$
    – toypajme
    Aug 29, 2013 at 23:44
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$\frac{f'(\frac{1}{n})}{f(\frac{1}{n})} = \frac{g'(\frac{1}{n})}{g(\frac{1}{n})}$ for $n=1,2, ... ,$ implies $f'(\frac{1}{n})g(\frac{1}{n})-g'(\frac{1}{n})f(\frac{1}{n})=0$ for $n=1,2, ... ,$. This implies that $\frac{f'(\frac{1}{n})g(\frac{1}{n})-g'(\frac{1}{n})f(\frac{1}{n})}{g(\frac{1}{n})^2}=\left(\frac{f(\frac{1}{n})}{g(\frac{1}{n})}\right)'=0$ and in turn that, since $f/g$, $(f/g)'$ are analytic and the latter agrees with the function $0$ on a bounded, convergent series of points, $(f/g)'=0$ on $A$, by the identity theorem and $f/g=c$ (on the set $A$) for some complex constant $c$.

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