I have a function $$f(z)=(z-1)^{3/5}(z+1)^{2/5}$$ and I have the branch of this function chosen such that $$-\pi<\arg(z\pm1)\leq\pi$$ How do I show that a branch cut is not required on the section $(-\infty,1)$ of the real axis?

I have defined $(z-1)=r_1e^{i\theta_1}$ and $(z+1)=r_2e^{i\theta_2}$ and thus can get $f(z)$ into the form $$r_1^{3/5}r_2^{2/5}e^{i/5(3\theta_1+2\theta_2)}$$ Now here is where I get confused but I think I need to find $f(x\pm0i)$on the section $(\infty,1)$ of the real axis and determine continuity but I don't know how to do this for this specific example and would appreciate some help. Thank you.


2 Answers 2


You only need to choose a branch cut so that all of the branch points lie on the cut. Remember that a branch point is a point where the function is discontinuous when traversing an arbitrarily small circuit about the point. By choosing a branch cut like this you're essentially acknowledging this behavior and choosing specifically where this discontinuity will occur.

The points $z = \pm 1$ are clearly branch points, and they are the only finite branch points. The point at infinity may also be a branch point. In this case it isn't (shown below), but if it were then you would need to choose your cut so that it contains all three branch points ($1$, $-1$, and $\infty$). You could do this by choosing it to be the real interval $(-\infty,1]$, but there are infinitely many other choices you could make too.

To see that the point at infinity is not a branch point, rewrite the function as

$$ f(z) = z\,\left(\frac{1}{z}-1\right)^{3/5}\left(\frac{1}{z}+1\right)^{2/5}. $$

Let $|z| > 1$. By doing this, we can ensure that neither of the two quantities $a = \frac{1}{z}-1$ or $b = \frac{1}{z}+1$ will make a circuit about the origin, regardless of what $z$ is doing. They are both restricted to unit disks centered at $-1$ and $1$, respectively. Indeed, $|a+1|<1$ and $|b-1|<1$.

Now, with this condition on $|z|$, if $z$ makes a circuit about the origin (equivalently, a circuit about infinity), the change in the argument of $f$ is precisely $2\pi$, and the function remains unchanged. We have thus continuously traversed an arbitrarily small circuit about infinity, and hence infinity is not a branch point of $f$.

As a result, we only need to choose a branch cut which includes $1$ and $-1$. The real inverval $[-1,1]$ is the simplest choice.

  • $\begingroup$ The residue at $\infty$ is the negative of the total of all the residues at the finite singularities. This unifies our answers. $\endgroup$
    – robjohn
    Jan 5, 2013 at 1:10

One only needs a branch cut between $-1$ and $1$. We can define $$ \log\left(\frac{w-1}{w+1}\right)=\frac{\pi i}{2}+\int_i^w\left(\frac1{z-1}-\frac1{z+1}\right)\,\mathrm{d}z $$ where the path from $i$ to $w$ does not cross the line segment from $-1$ to $1$. This forces the difference of any two paths from $i$ to $w$ to circle both singularities the same number of times. Since their residues are $1$ and $-1$, the total residue would be $0$ and so the function computed using either path would be the same.

Therefore, we can define $$ (z-1)^{3/5}(z+1)^{2/5}=\left(\frac{z-1}{z+1}\right)^{3/5}(z+1) $$ using the same branch cut.


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