Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer
    
The residue at $\infty$ is the negative of the total of all the residues at the finite singularities. This unifies our answers. –  robjohn Jan 5 '13 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.