Notes - Complex Analysis MT23, Multifunctions and branch cuts

Created: November 02, 2023

Flashcards

Can you define a multifunction on $U \subseteq \mathbb C$?

A map $f : U \to \mathcal P(\mathbb C)$.

Suppose $f : U \to \mathcal P(\mathbb C)$ is a multifunction. Then what is a branch $g : V \to \mathbb C$ where $V \subseteq U$?

Any function $g$ such that for all $z \in V$, $g(z) \in f(z)$.

How many solutions does $z^\alpha = w$ where $\alpha \in \mathbb Q$ and $z, w \in \mathbb C$ have?

$n$ inverses.

How many solutions does $z^\alpha = w$ where $\alpha \in \mathbb Q^C$ and $z, w \in \mathbb C$ have?

Infinitely many inverses.

How could you define a holomorphic branch $L(z)$ of $\log z$ on the cut plane $\mathbb C \setminus R _ \alpha$ such that $L(1) = 0$ and $R _ \alpha = \{z \in \mathbb C \mid z = re^{i\alpha}, r \in [0, \infty)\}$?

We can write each $z$ on the cut plane with $z = re^{i\theta}$ with $r > 0$ and $\alpha - 2\pi < \phi < \alpha$. Then define $L(z) = \log r + i\theta$.

How could you define a holomorphic branch of $\text{Log}(z^2 - 1)$ on $\mathbb C \setminus ((-\infty, -1] \cup [1, \infty))$?

There are two approaches:

  • Note that $z \mapsto z^2 - 1$ maps $\mathbb C \setminus ((-\infty, -1] \cup [1, \infty))$ to $\mathbb C \setminus [0, \infty)$ and then just define a branch of the logarithm on this region and take the composition, or
  • Define $z = 1 = r _ 1 e^{i\theta _ 1}$ and $z + 1 = r _ 2 e^{i\theta _ 2}$ where $0 < \theta _ 1 < 2\pi$ and $-\pi < \theta _ 2 < \pi$. Then define $L(z^2 -1) = \log(r _ 1 r _ 2) + i(\theta _ 1 + \theta _ 2)$ gives the required branch.

How could you justify that e.g.

\[L(z^2 -1) = \log(r_1 r_2) + i(\theta_1 + \theta_2)\]

defined on $\mathbb C \setminus ((-\infty, -1] \cup [1, \infty))$ is single-valued as you cross the real axis away from the cut?

Note that both $\theta _ 1$ and $\theta _ 2$ vary continuously across the cut.

Suppose:

  • $f : U \to \mathcal P(\mathbb C)$ is a multi-valued function
  • $U$ is an open subset of $\mathbb C$

What does it mean for a point $z _ 0$ to be a branch point of $f$?

$z _ 0$ is not a branch point if there is an open disc $D \subseteq U$ containing $z _ 0$ such that there is a holomorphic branch of $f$ defined on $D \setminus \{z _ 0\}$ .

$z _ 0$ is a branch point otherwise.



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