# Is there a holomorphic function such that $(f(z))^3=z-z^2$

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!
• 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? – ryszard eggink Feb 1 at 17:42
• @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? – 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.