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I have a problem with the following exercise

Find all functions $f$ such that $f$ is a holomorphic function in $B(0,1)$ and $2f'(\frac{1}{n})f(\frac{1}{n})=1$, $n=2,3,4,\dots$

I deduced that $g(z):=f'(z)f(z)$ must be equal to $\frac{1}{2}$ in $B(0,1)$ (from Identity Theorem). But I don't know what I should do next... Could you give me some advice? (which theorem should I use or sth like that) I would be very grateful. Ive spent 1.5 h on it already...

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    $\begingroup$ Note $[f(z)^2 - z]' = 0$. $\endgroup$ – kobe Jan 10 '15 at 16:47
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Here's a place to start: if you were dealing with that equation on $(-1,1)$, you could integrate both sides from $x=0$ to $x=y$, getting

$$f(y)^2-f(0)^2=y$$

So if we define $C=f(0)^2$ then the solution set is $f(y)=\pm \sqrt{y+C}$.

When can you extend this answer to the disk to get a holomorphic function?

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  • $\begingroup$ It will be that for any $|z|<1$ we have $f(z)^2-f(0)^2=z$ ? (from fundamental theorem of calculus (complex version for curve integrals)? $\endgroup$ – luka5z Jan 10 '15 at 17:06
  • $\begingroup$ I need to find such $C$ that square root will be holomorphic function in B(0,1). Give me a while please $\endgroup$ – luka5z Jan 10 '15 at 17:21
  • $\begingroup$ So $Re(C)\not\in (-\infty,1]$ or $|Im(C)|\not\le 1$ $\endgroup$ – luka5z Jan 10 '15 at 17:29
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    $\begingroup$ I think you should find $|C| \geq 1$ is enough for some branch of the square root to work. The principal one might not, but that doesn't matter. Ultimately, since $y+C$ is a biholomorphism (holomorphic with holomorphic inverse), the key is that $y+C$ is never $0$, since $0$ is the branch point of the square root Riemann surface. $\endgroup$ – Ian Jan 10 '15 at 17:41

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