Is there a holomorphic function $f:C-[0,1]$ such that $(f(z))^3=z-z^2$ for all $z\in C-[0,1]$

My intuition tells me that not really, for instance $$(z-z^2)^{\frac{1}{3}}$$

does not have a unique branch on this set, but I do not know how to formally prove it.


The function $z\to z-z^2$ has a double pole at infinity, which means that your $f$ would have a pole of order $2/3$ there!

  • $\begingroup$ which means that we can write $f=z^{\frac{2}{3}}g(z)$ where g is holomorphic and g(infty) is not equal to ininity. which would mean that $\frac{f}{g}$ would be holomorphic but it is impossible, since $z^{\frac{2}{3}}$ is not holomorphic in infinity? $\endgroup$ – ryszard eggink Feb 1 at 17:42
  • $\begingroup$ @ryszardeggink or consider $h\colon z\mapsto \frac1{(1/z)+(1/z)^2}$ on $\Bbb C-(\{0\}\cup [1,\infty))$. It has a removable singularity at $0$ and power series $h(z)=z^2+\ldots$. But what should the power sereis of $z\mapsto \frac1{f(1/z)}$ look like? $\endgroup$ – Hagen von Eitzen Feb 1 at 19:26

The zeroes of $f$ must be the same as those of $z-z^2$, namely $0$ and $1$.

What is the winding number of $f(z)$ as $z$ describes a small circle around $0$? Since $z-z^2$ winds around $0$ once, you need an integer that gives $1$ when multiplied by $3$. But that is impossible.


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