I want to find the number of analytic functions on the unit disk $\mathbb{E} = \{z:|z|<1\}$ such that $f:\mathbb{E}\ \rightarrow \mathbb{C}$ is analytic and $(f(\frac{1}{n}))^3=\frac{n}{n+1}$ for every $n \in \mathbb{N}$.
I thought of doing the following:
We know that $\frac{1}{n}\rightarrow 0$ as $n \rightarrow \infty$, so $\mathbb{E}$ contains an accumulation point of any function $f$. The identity theorem then tells us that there is at most one analytic function $f$ such that the criterium holds.
The problem now, is that we don't have an explicit statement for $f(\frac{1}{n})$, rather we have a statement for $(f(\frac{1}{n}))^3$. I thought that we could just take the cube root to get an explicit statement for $f(\frac{1}{n})= \sqrt[3]{\frac{n}{n+1}}$. The function $g(z) $ that we can make by extending $f(\frac{1}{n})$, would be $g(z)=\sqrt[3]{\frac{1}{z+1}}$ (where we take $z \mapsto \sqrt[3]{z} = \exp\{\frac{1}{3}(\ln(|z|)+i\cdot \text{Arg}(z))\}$. Since this is analytic on our disk, we know that there is just one analytic function such that $(f(\frac{1}{n}))^3=\frac{n}{n+1}$ for every $n \in \mathbb{N}$.
My problem with this method however, is that I don't know whether $f(\frac{1}{n})= \sqrt[3]{\frac{n}{n+1}}$ is the only function for $f$ that satisfies $(f(\frac{1}{n}))^3=\frac{n}{n+1}$. It feels weird to assume that there is just one answer to this function. My question is now, is my suspicion grounded?
What would we do if we had $(f(\frac{1}{n}))^k=\frac{n}{n+1}$ for $k \in \mathbb{N} \backslash\{1\}$? I know that $k=2$ gives two analytic functions, Does this extend to the following: For every even $k$ we have 2 answers, for every odd $k$ we have 1?
Thanks for your time,
K. Kamal
Best Answer
YES it is. In fact, there are, for any nonzero complex number $w$, exactly $k$ solutions to the equation $z^k=w$. This is because you can multiply any solution $z_1$ by $e^{\frac{2\pi}{k}i}$ (that is, rotate the solution around $0$ by an angle of $\frac{2\pi}{k}$) and still get a solution.
In your case, you could also have $$e^{\frac13(\ln|z| + i(\mathrm{Arg}(z) + \frac{1\cdot 2\pi}{3})}$$ or $$e^{\frac13(\ln|z| + i(\mathrm{Arg}(z) + \frac{2\cdot 2\pi}{3})}$$
and still get a solution.