# Branch cut of $w=\left(\dfrac{z+1}{z-1}\right)^{1/3}$

I'd like todetermine a branch cut for the function $w=\left(\dfrac{z+1}{z-1}\right)^{1/3}$ that allows to construct analytic branches defined on $|z|>1 \;,\; \forall z\in \mathbb C$. How can I do this?

I noticed: branch points are $z=1$ and $z=-1$. So I'd say: $[-1,1]$ is the branch cut needed. Is this correct?

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–  Did Oct 30 '13 at 10:43

Note that the quotient $$\frac{z+1}{z-1}$$ is real and nonpositive only on the interval $[-1,1]$. Thus , it is possible to define an analytic branch of $$\log \left( \frac{z+1}{z-1} \right)$$ in $\mathbb C \setminus [-1,1]$ where the imaginary part lies in $(-\pi,\pi)$.
Then $$\exp \left( \frac{1}{3} \log \left(\frac{z+1}{z-1} \right) \right)$$ provides the desired analytic branch.
@Bob How do you define the complex powers $z^w$? –  user1337 Oct 30 '13 at 10:31
In this case I'd say: $w^3= \frac{z+1}{z-1}$. Is that wrong? –  Bob Oct 30 '13 at 10:32
@Bob for any $z \neq -1$ there are three different possible candidates for $w$. Using a definite branch of the logarithm picks one. –  user1337 Oct 30 '13 at 10:35