# $(z-1)^a(z+1)^b$ has an analytical branch in $\mathbb{C}-[-1,1]$ iff $a+b \in \mathbb{Z}$

Prove that $(z-1)^a(z+1)^b$ has an analytical branch in $\mathbb{C}-[-1,1]$ iff $a+b \in \mathbb{Z}$.

I think I proved the "if" part: Let us define the branch $$f(z)=\bigg( \frac{z-1}{z+1}\bigg)^a(z+1)^{a+b}$$

But I have no idea how to prove the "only if" part. Any ideas?

Thanks!

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Since you are allowed to multiply the function with $(z-1)^n(z+1)^m$ with $n,m\in\mathbb N$, we may assume that $a,b>1$. Consider $z=re^{i\phi}$ with $r$ big. Then $(z-1)^a(z+1)^b= r^{a+b}e^{(a+b)i\phi}\left(1+O(\frac1r)\right)+C$. For this to "close" if we fix big $r$ and let $\phi$ change from $0$ to $2\pi$, we need that $(a+b)$ is an integer. More precisely, we have \begin{align}0&=r^{a+b}e^{(a+b)i\cdot2\pi}\left(1+O(1/r)\right)-r^{a+b}e^{(a+b)i\cdot0}\left(1+O(1/r)\right)\\ &=r^{a+b}(e^{(a+b)i\cdot2\pi}-1+O(1/r)) \end{align} i.e. $$e^{(a+b)i\cdot2\pi}-1=O(1/r),$$ hence $e^{(a+b)i\cdot2\pi}=1$.