# Holomorphic functions on unit disc

Let $f,g$ be holomorphic on $\mathbb{D}:=\lbrace z\in\mathbb{C}:|z|<1\rbrace$, $f\neq0,g\neq0$, such that $$\frac{f^{\prime}}{f}(\frac{1}{n})=\frac{g^{\prime}}{g}(\frac{1}{n})$$ for all natural $n\geq1$. Does it imply that $f=Cg$, where $C$ is some constant?

Let $A:=\lbrace\frac{1}{n}:n\geq1\rbrace$ and $h:=\frac{f^{\prime}}{f}-\frac{g^{\prime}}{g}$. Now, $h$ is holomprphic on $\mathbb{D}$ and disappears on a subset of $\mathbb{D}$ which has a limit point. Thus $h=0$, so $\frac{f^{\prime}}{f}=\frac{g^{\prime}}{g}$ on $\mathbb{D}$.

Could someone help with the next steps? Or maybe $f$ doesn't have to be in the form described above?

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Notice that $f'/f \equiv \operatorname{d}\log f$. Moreover, if $f$ and $g$ are holomorphic then they are continuous and so when you apply a function to a sequence, the limit of the values if the value of the limit. (If $x_k \to \ell$ then $f(x_k) \to f(\ell)$.) –  Fly by Night Jan 31 '13 at 19:20

Notice that your last statement is equivalent to ${f'g-g'f}=0$, since $f,g \neq 0$. Now there is no harm in dividing that expression by $g^2$. You get ${f'g-g'f}{g^2}=0$. So, $(f/g)'=0$ and so your result follows.

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The "it follows" bit is what the OP was asking for help with. One needs to argue in terms of convergent sequences, limit points, and continuous functions. (We're not told that $f/f' \equiv g/g'$, just that they agree on a convergent sequence.) –  Fly by Night Jan 31 '13 at 19:55
@FlybyNight But it appears that the O.P figured out how $f/f'=g/g'$ on the entire disk from what the problem told. (See his last but third sentence). I think what the OP needed was to see that $f/f'=g/g'$ is really the quotient rule in disguise. The rest should not be too difficult. Perhaps I am wrong. Since $f,g$ are holomorphic and neither is zero in our domain, $f/g$ is defined and holomorphic. Since it's derivative is zero, it is a constant. –  Galois Group Feb 1 '13 at 21:50

If $f$ and $g$ are allowed to be $0$ somewhere (but not everywhere), the result is still true. Suppose $g(z) = z^k G(z)$ where $k$ is a nonnegative integer and $G$ is holomorphic on ${\mathbb D}$ with $G(0) \ne 0$. Then $$\dfrac{g'(z)}{g(z)} = \dfrac{k}{z} + \dfrac{G'(z)}{G(z)} = \dfrac{k}{z} + O(1) \text{ as } z \to 0$$ So from the sequence $(g'/g)(1/n)$ we can determine $k$. Thus if $(g'/g)(1/n) = (f'/f)(1/n)$ for all $n$ we also have $f(z) = z^k F(z)$ (for the same $k$) where $F$ is holomorphic on ${\mathbb D}$ with $F(0) \ne 0$, and $F'/F = G'/G$. Since you already proved the result for functions nonzero at $0$, it follows that $f/g = F/G$ is constant.

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Hint: Since $$\frac{f'}{f}=\frac{g'}{g}$$ on $\Bbb D$, then $f'g-fg'=0$ on $\Bbb D$, and so $$\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}=0$$ on $\Bbb D$. Can you get the rest of the way from there and see why $\frac{f}{g}$ is holomorphic on $\Bbb D$?

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That's not the quotient rule for derivatives. Should be $g^2$ on the bottom. –  Christopher A. Wong Jan 31 '13 at 19:26
Thanks, @Christopher. –  Cameron Buie Jan 31 '13 at 19:30