Complex Analysis: Find all analytic functions such that $(f(\frac{1}{p}))^3=\frac{p+1}{p}$ I've started the following problem but I'm stuck now.
Find all analytic functions $f:\mathbb{E}\to\mathbb{C}$ that satisfy
$(f(\frac{1}{p}))^3=\frac{p+1}{p}\qquad$ for all natural numbers $p$ and $\mathbb{E}$ being the unitdisk $\{z:|z|<1\}$.
My answer so far:
I've found the following functions that satisfy the condition:
$f_1(z)=(1+z)^\frac{1}{3},\quad f_2(z)=-(-1)^\frac{1}{3}(1+z)^\frac{1}{3},\quad f_3=(-1)^\frac{2}{3}(1+z)^\frac{1}{3}$.
But I fail to show how these functions are analytic on the unitdisk and how I would have to define them on the unitdisk. Also, do I have to prove there is no distinct $f_4$?
 A: $(f(\frac{1}{p}))^3=\frac{p+1}{p}$ is equivalent to  $f(\frac{1}{p})$ being either one of the three cube roots of $\frac{p+1}{p}$. 
Note that there is an analytic cube root of $1+z$ when $z\in \Bbb E$ because there is an analytic logarithm defined on $1+\Bbb E$: indeed, $1+\Bbb E$ does not contain $0$ or a loop that goes around it. You can then freely define $$f_1(z)=e^{\frac13 \log(1+z)}$$
with $\log(1+z)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}z^n$ being analytic on $\Bbb E$.
Now as was said in the beginning and pointed out by Martin R, there are three cube roots of any non-zero complex number, which yields two other solutions $$f_2(z)=e^{\frac{2i\pi}{3}}f_1(z)\qquad\text{and}\qquad f_3(z)=e^{\frac{4i\pi}{3}}f_1(z).$$

Proof that these are the only three solutions.
We know that for each $z=\frac 1p$, $f(z)$ coincides with either $f_1(z)$, $f_2(z)$ or $f_3(z)$. Why couldn't it be a different $f_i$ for each $p$? Because by the pigeonhole principle, there is at least one $f_i$ that coincides with $f$ on infinitely many $\frac 1p$'s, therefore on a set with a limit point, and you are done by analytic continuation. 

See On the Power Series for $\log(1+z)$, published in the Annals of Mathematics in 1916!
