Notes - Complex Analysis MT23, Winding numbers
Flashcards
We want to be able to define the winding number of some path $\gamma : [0, 1] \to \mathbb C \setminus \{0\}$ as
\[\arg\gamma(b) - \arg\gamma(a)\]
but this doesn’t work because $\arg$ is not a continuous function. Can you state a theorem that lets us rewrite $\gamma$ in a way that lets us do something similar?
Let $\gamma$ be such a path. Then $\exists a : [0, 1] \to \mathbb R$ such that:
- $a$ is continuous
- $\gamma(t) = \vert \gamma(t) \vert e^{2\pi i a(t)}$
Moreover, if $a$ is unique up to adding or subtracting a natural number.
Can you define, in terms of some function $a(t)$, the winding number of a closed path $\gamma : [0, 1] \to \mathbb C \setminus \{0\}$, denoted $I(\gamma, 0)$?
Write
\[\gamma(t) = |\gamma(t)|e^{2\pi i a(t)}\]Then
\[I(\gamma, 0) = a(1) - a(0)\]For some closed path $\gamma : [0, 1] \to \mathbb C \setminus \{0\}$ where $\gamma(t) = \vert \gamma(t) \vert e^{2\pi i a(t)}$, the winding number is $I(\gamma, 0) = a(1) - a(0)$. How is the winding number $I(\gamma, z _ 0)$ defined for points $z _ 0$ that aren’t the origin?
where
\[t(z) = z - z_0\]Suppose
- $\gamma : [a, b] \to \mathbb C$ is a piecewise $C^1$ closed path
- $z _ 0 \in \mathbb C$ but $z _ 0 \notin \gamma^\star$.
Then how can we equivalently write the winding number $I(\gamma, z _ 0)$ as an integral, so that we are not relying on rewriting $\gamma(t)$ as $ \vert \gamma(t) \vert e^{2\pi i a(t)}$?
Quickly prove that if
- $\gamma : [a, b] \to \mathbb C$ is a piecewise $C^1$ closed path
- $z _ 0 \in \mathbb C$ but $z _ 0 \notin \gamma^\star$
Then
\[I(\gamma, z_0) = \frac{1}{2\pi i} \int_\gamma \frac{1}{z-z_0} \text dz\]
Wlog $\gamma : [0, 1] \to \mathbb C$ and $\gamma(t) = z _ 0 + r(t)e^{2\pi a(t)}$. Then
\[\begin{aligned} \int_\gamma \frac{1}{z-z_0} \text dz &= \int^1_0 \frac{1}{r(t)e^{2\pi ia(t)}\\,}(r'(t) + 2\pi ir(t) a'(t))e^{2\pi i a(t)} \text dz \\\\ &= \int^1_0 \frac{r'(t)}{r(t)} + 2\pi i a'(t) \text dt \\\\ &= \left[ \log(r(t)) - 2\pi i a(t) \right]^1_0 \\\\ &= 2\pi i(a(1) - a(0)) \end{aligned}\]Let $U$ be an open set in $\mathbb C$ and let $\gamma : [0, 1] \to U$ be a closed path. Can you define $I _ f(\gamma, w)$ where $f(z)$ is a continuous function on $\gamma^\star$?
For $U$ open, $\gamma$ path, and $f$ continuous function on $\gamma^\star$, $I _ f(\gamma, w)$ is given by
\[I_f(\gamma, w) = \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z-w} \text dz\]
What useful property does this function possess?
It is analytic in $w$.
For $U$ open, $\gamma$ path, and $f$ continuous function on $\gamma^\star$, $I _ f(\gamma, w)$ is given by
\[I_f(\gamma, w) = \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z-w} \text dz\]
This function is analytic in $w$. Using $f$ given by $x \mapsto 1$, what can we deduce about the winding numbers of a path on the connected components of $\mathbb C \setminus \gamma^\star$?
Analytic, so continuous, and since integer valued (this is just the winding number), it is constant on the connected components of $\mathbb C \setminus \gamma^\star$.
Let $\gamma : [0, 1] \to \mathbb C$ be a closed path. What does it mean for $z$ to be on the inside of $\gamma$?
$z \notin \gamma^\star$ and $I(\gamma, z) \ne 0$.
What does it mean for a path $\gamma : [0, 1] \to \mathbb C$ to be simple?
Can you state the Jordan curve theorem?
If
- $\gamma : [0, 1] \to \mathbb C$ is a simple closed curve
Then:
- $\mathbb C \setminus \gamma^\star$ is the union of one bounded and unbounded connected component
- On the bounded component, $I(\gamma, z)$ is either $1$ or $-1$.
What does it mean for a simple closed curve to be positively or negatively oriented?
For $z$ on the inside of $\gamma$,
- $I(\gamma, z) = 1$
- $I(\gamma, z) = -1$
Suppose:
- $U \subseteq \mathbb C$ open
- $\gamma : [0, 1] \to U$ closed path
- $f : U \to \mathbb C$ continuous on $\gamma^\star$
and
\[I_f(\gamma, w) := \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z - w} \text dz\]
Quickly prove:
- $I _ f(\gamma, w)$ is analytic in $w$
- The winding number $I(\gamma, w)$ is constant on the connected components of $\mathbb C \setminus \gamma^\star$
- There is a nice expression integral for the derivatives of $g(w) := I _ f(\gamma, w)$ which you should state (this, along with Cauchy’s integral formula, gives a nice expression for the derivatives of an arbitrary holomorphic function)
We want to show that given $z _ 0 \notin \gamma^\star$, we can find a disk $B(z _ 0, r)$ within which $I _ f(\gamma, w)$ is given by a powers series in $(w - z _ 0)$. Translating, we can assume $z _ 0 = 0$.
Since $\mathbb C \setminus \gamma^\star$ is open, $\exists r > 0$ such that $B(0, 2r) \cap \gamma^\star = \emptyset$. Now note if $w \in B(0, r)$ and $z \in \gamma^\star$, $ \vert w/z \vert < 1/2$. Furthermore, since $\gamma^\star$ is compact,
\[M = \sup \\{|f(z)| : z\in \gamma^\star\\}\]is finite, and so
\[\begin{aligned} \left| f(z) \cdot \frac{w^n}{z^{n+1}\\,} \right| &= |f(z)| \cdot |z|^{-1} \cdot \left|\frac w z\right|^n \\\\ &< M \cdot (2r)^{-1 } \cdot (1/2)^n \end{aligned}\]Then
\[\begin{aligned} \frac{f(z)}{z - w} &= \frac{f(z)}{z} \frac{1}{1-\frac w z} \\\\ &= \sum^\infty_{n = 0} \frac{f(z)}{z} \left(\frac w z\right)^n \\\\ &= \sum^\infty_{n = 0} \frac{f(z)}{z^{n+1}\\,} w^n \end{aligned}\]where in the second equality we have used Weierstrass $M$-test, and the bound we found above. Then the series converges uniformly on $\gamma^\star$ to $f(z) / (z-w)$, so $\forall w \in B(0, r)$ we have
\[\begin{aligned} I_f(\gamma, w) &= \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z-w} \text dz \\\\ &= \sum^\infty_{n=0} \left(\frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z^{n+1}\\,}\text dz\right) w^n \end{aligned}\]So $I _ f(\gamma, w)$ is given as a convergent power series in $B(0, r)$.
If $f = 1$, then $I _ 1(\gamma, z) = I(\gamma, z)$. Since the winding number is integer valued, and is continuous on $\mathbb C \setminus \gamma^\star$, it must be constant on the connected components.
Since the power series of $I _ f(\gamma, w)$ determines its derivatives, we must have
\[\frac{\text d^n}{\text dw^n} I_f(\gamma, w) = \frac{n!}{2\pi i} \int_\gamma \frac{f(z)}{z^{n+1}\\,} \text dz\]Proofs
Prove that if $\gamma : [0, 1] \to \mathbb C$ is a path, then there is a function $a : [0, 1] \to \mathbb R$ such that
- $a$ is continuous
- $\gamma(t) = \vert \gamma(t) \vert e^{2\pi i a(t)}$
Moreover, if $a$ is unique up to adding or subtracting a natural number.
Todo, can replace $\gamma(t)$ by $\gamma(t)/ \vert \gamma(t) \vert $.