I have the following function: $$f(z) = \left( \frac{z}{z+x_0} \right)^q,$$ where $q \in \mathbb Q, \ q = \frac{p}{r}$ with $p \neq r$ and $r \neq 0$ and $x_0 \in \mathbb R$, $x_0 > 0$.
I want to show that $z = 0$ is a branch point.
My attempt: Calculate the contour integral $\oint f(z)dz$ on a loop around $z = 0$. If we get the same value for the starting point and endpoint, the integral should vanish. But, as we want to show that $z = 0$ is a branch point, the integral shouldn't vanish.
Let our starting point be $a = 0$ and our endpoint $b = 2\pi$. We rewrite $z$ as $z(\theta) = re^{i\theta}$, where $r$ is the radius, i.e. the absolute value of $z$ and $\theta$ is the argument of $z$, i.e. $arg(z)$. We then calculate $z'(\theta) = rie^{i\theta}$ and let $r << 1$ and $0 \leq \theta \leq 2\pi$. Then we have:
$$ \int_{\gamma} f(z)dz = \int_{a}^{b} f(z(\theta)) z'(\theta)d\theta = \int_{0}^{2\pi} \left( \frac{re^{i\theta}}{re^{i\theta}+x_0} \right)^{q} r i e^{i\theta}d\theta \approx \int_{0}^{2\pi} \left( \frac{re^{i\theta}}{x_0} \right)^{q} r i e^{i\theta}d\theta \\ = \frac{ir^{q+1}}{x_0^{q}} \int_{0}^{2\pi} e^{i\theta(q+1)} d\theta = \frac{ir^{q+1}}{x_0^{q}} \frac{1}{i(q+1)} \left( e^{2\pi(q+1)i} – e^{0(q+1)i} \right) = \frac{ir^{q+1}}{x_0^{q}} \frac{1}{i(q+1)} \left( e^{2\pi(q+1)i} – 1 \right)
.$$
In the $4$th step I approximate $\left({re^{i\theta}+x_0}\right)$ in the denominator as $x_0$ because $r << 1$. The reasoning why the integral is not $0$ is that $e^{2\pi(q+1)i} – 1 \neq 0$ for any $q$ as defined above.
Is the approximation that I made a valid step to made or can this be shown more rigorous?
Best Answer
I think this is a good exercise for me to review the branch cuts and points. To solve this question, I would prefer to start from basic definitions because it's easy to understand.
The function $f(z)$ can be written as $\sqrt[r]{z^{p}(z+x_0)^{-p}}$. Now let the first $z=r_{1}e^{i\theta_{1}}$ and the second $z+x_{0}=z-(-x_{0})=r_{2}e^{i\theta_{2}}$, so it turns to $\sqrt[r]{r_{1}^{p}r_{2}^{-p}e^{ip\theta_{1}-ip\theta_{2}}}$, by then we turn it into a geometry problem.
Consider the first counterclockwise circle centered at $z=0$, only the origin point inside. Let a point on the circle starts from $\theta_{1}=0$ and goes around one whole circle, the argument change is $+2\pi$ while the argument of $z=-x_{0}$ remains same, look back the polar form of $f(z)$ we shall find it becomes $$\sqrt[r]{r_{1}^{p}r_{2}^{-p}e^{ip(\theta_{1}+2\pi)-ip\theta_{2}}}=\sqrt[r]{r_{1}^{p}r_{2}^{-p}e^{ip\theta_{1}-ip\theta_{2}}}e^{i\frac{2p\pi}{r}}=\sqrt[r]{r_{1}^{p}r_{2}^{-p}e^{ip\theta_{1}-ip\theta_{2}}}e^{i(2q\pi)}$$
You shall find when $q=\frac{p}{r}=k$ (here $k$ is an integer, it must be considered an integer is also a rational number), by $e^{2k\pi i}=1(k\in\mathbb{Z})$ the function value stays same. When $q$ is not an integer, $z=0$ is indeed a branch point.
BTW You can also do similar analysis to $z=-x_{0}$.