Understanding the function $f(z) = \left(\frac{z}{z + x_0} \right)^{\kappa}$

Question

I have the following complex function: $$f(z) = \left(\frac{z}{z + x_0} \right)^{\kappa}$$

where $x_0 \in \mathbb R$ and $x_0 > 0$. $\kappa$ is a parameter for which we look at three cases:

1. case: $\kappa \in \mathbb N$ and $\kappa \neq 1$
2. case: $\kappa \in \mathbb Q$
3. case: $\kappa \in \mathbb I$, i.e. $\kappa$ is irrational

My professor told me that for the first two cases, we get a finite number of Riemann sheets for $f(z)$. But in the 3. case we get an infinite number of Riemann sheets.

My understanding/assumptions regarding the first two cases: First, by Riemann sheets I mean branches of a multivalued function.

If $\kappa$ is rational (which also includes the $\kappa \in \mathbb N$ case), it has the form $\kappa = \frac{m}{n}$ with $n \neq 0$. Therefore, we can rewrite $f(z)$ as $f(z) = \frac{z^{\frac{m}{n}}}{(z + x_0)^{\frac{m}{n}}} = \frac{(z^m)^{\frac{1}{n}}}{((z + x_0)^m)^{\frac{1}{n}}}$. As far as I understand it, for both $(z^m)^{\frac{1}{n}}$ and $((z + x_0)^m)^{\frac{1}{n}}$, the $\frac{1}{n}$ "part" is what causes the branches, since this represents the $n$-th root and we know that any function $g(z) = z^{\frac{1}{n}}$ has exactly $n$ branches. Therefore, the number of branches is finite.

But why is it that when $\kappa$ is irrational, the number of branches becomes infinite?

And this function somehow "more known" in mathematics and/or physics and thus more studied? Does it have a name?

I somehow understood my professor that it appears in the definition of a propagator in Quantum Field Theory, but I'm not entirely sure.

And if I made a mistake somewhere in my understanding, please feel free to correct me.

P.S. is there a more formal proof (or explanation) which shows why for rational $\kappa$ the number of Riemann sheets is finite but for irrational it's infinite?

Answer 1

$\DeclareMathOperator{\Log}{Log}\DeclareMathOperator{\im}{Im}\DeclareMathOperator{\pow}{\texttt{powC}}$Let $\kappa$ be a real number, let $\Log$ denote the principal branch of logarithm (satisfying $-\pi < \im\Log \leq \pi$), and let $\log$ denote the "total logarithm", which assigns to a non-zero complex number $w$ the set of all complex numbers $u$ satisfying $w = \exp u$. In other words, if $w \neq 0$, then \begin{align*} \log w &= \{\Log w + 2\pi ni : \text{$n$ an integer}\} \\ &= (\Log w) + 2\pi i \mathbf{Z}. \end{align*}

Define the "total $\kappa$th power function," formally $z^{\kappa}$ but less freighted with expectations about rules of exponents, by \begin{align*} \pow(w, \kappa) &= \exp(\kappa\log w) \\ &= \{\exp(\kappa\Log w + 2\pi n\kappa i) : \text{$n$ an integer}\} \\ &= \{\exp(\kappa\Log w) \cdot \exp(2\pi n\kappa i) : \text{$n$ an integer}\}. \end{align*} The factor $\exp(\kappa\Log w)$ is well-defined because $\Log$ is single-valued (despite the discontinuity on the non-positive real axis). The multi-valuedness, if any, therefore arises from the set $\{\exp(2\pi n\kappa i) : \text{$n$ an integer}\}$.

Because $\exp(2\pi n\kappa i) = 1$ if and only if $n\kappa$ is an integer, we deduce:

• If $\kappa$ is an integer, then $\exp(2\pi n\kappa i) = 1$ for every integer $n$, and $\pow(\ , \kappa)$ is single-valued.

• If $\kappa = p/q$ is rational in lowest terms and $1 < q$, then $$ \exp(2\pi n\kappa i) = \exp(2\pi npi/q) $$ takes precisely $q$ distinct values as $n$ runs over the integers, so $\pow(\ , \kappa)$ is $q$-valued.

• If $\kappa$ is irrational, then $\exp(2\pi n\kappa i) = 1$ if and only if $n\kappa$ is an integer, if and only if $n = 0$. Consequently, if $n$ and $n'$ are integers, then $\exp(2\pi n\kappa i) = \exp(2\pi n'\kappa i)$ if and only if $$ 1 = \frac{\exp(2\pi n\kappa i)}{\exp(2\pi n'\kappa i)} = \exp\bigl(2\pi (n - n')\kappa i\bigr), $$ if and only if $n = n'$. We conclude $\pow(\ , \kappa)$ takes infinitely many distinct values, one for each integer $n$.

<>

So what about the function $f$ in question? The conceptual point is, $w := z/(z + x_{0})$ is an injective function of $z$ (given by a Möbius transformation), so the multi-valuedness of $f$ mirrors multi-valuedness of $\pow(\ , \kappa)$ with two exceptions:

• If $z = 0$ then $w = 0$ and $f$ is undefined.

• If $z = -x_{0}$, then $w = \infty$ (or is undefined, depending if we're working in the Riemann sphere or the complex plane), and again $f$ is undefined.