Notes - Complex Analysis MT23, Argument principle

Created: December 14, 2023

Flashcards

Suppose:

  • $f : U \to \mathbb C$
  • $f$ meromorphic
  • $f$ has a zero or pole of order $k$ at $z _ 0$

Then quickly prove that you can construct a new function that has a simple pole at $z _ 0$ with residue of $k$ or $-k$.

Suppose we are in the case where $f$ has a zero of order $k$. Consider

\[\frac{f'(z)}{f(z)}\]

Then

\[f(z) = (z-z_0)^k g(z)\]

where $g(z)$ is holomorphic near $z _ 0$ and $g(z _ 0) \ne 0$. Hence

\[\frac{f'(z)}{f(z)} = \frac{k}{z - z_0} + \frac{g'(z)}{g(z)}\]

and note that $g’(z) / g(z)$ is holomorphic near $z _ 0$, so it has a simple pole with residue $k$. The other case is similar (thanks notes!).

Can you state the argument principle about an meromorphic function $f : U \to \mathbb C$ on an open set?

Suppose

  • $U$ is an open set
  • $f : U \to \mathbb C$ meromorphic
  • $B(a, r) \subseteq U$
  • $N$ is the number of zeros of $f$ inside $B(a, r)$ counted with multiplicity
  • $P$ is the number of poles of $f$ inside $B(a, r)$ counted with multiplicity ins
  • $\gamma(t) = a + re^{2\pi i t}$

Then

\[\frac{1}{2\pi i} \int_\gamma \frac{f'(z)}{f(z)} \text dz = N - P\]

The argument principle states that if

  • $U$ is an open set
  • $f : U \to \mathbb C$ meromorphic
  • $B(a, r) \subseteq U$
  • $N$ is the number of zeros of $f$ inside $B(a, r)$ counted with multiplicity
  • $P$ is the number of poles of $f$ inside $B(a, r)$ counted with multiplicity
  • $\gamma(t) = a + re^{2\pi i t}$

then

\[\frac{1}{2\pi i} \int_\gamma \frac{f'(z)}{f(z)} \text dz = N - P\]

Quickly justify that this is in fact the same as the number of times the path $f \circ \gamma$ winds about the origin, i.e. $I _ {f}(\gamma, 0)$
\[\begin{aligned} I_f(\gamma, 0) &= \frac{1}{2\pi i} \int_{f \circ \gamma} \frac{1}{w} \text dw \\\\ &= \frac{1}{2\pi i} \int^1_0 \frac{1}{f(\gamma(t))} f'(\gamma(t))\gamma'(t) \text dt \\\\ &= \frac{1}{2\pi i} \int_\gamma \frac{f'(z)}{f(z)} \text d z \end{aligned}\]

Quickly prove the argument principle, i.e. that if

  • $U$ is an open set
  • $f : U \to \mathbb C$ meromorphic
  • $B(a, r) \subseteq U$
  • $N$ is the number of zeros of $f$ inside $B(a, r)$ counted with multiplicity
  • $P$ is the number of poles of $f$ inside $B(a, r)$ counted with multiplicity
  • $\gamma(t) = a + re^{2\pi i t}$

then

\[\frac{1}{2\pi i} \int_\gamma \frac{f'(z)}{f(z)} \text dz = N - P\]

Define

\[h(z) := \frac{f'(z)}{f(z)}\]

It’s nice to think of this as a “helper” function that helps us count the zeroes and poles of $f$. Suppose that $f$ has a zero of order $k$ at $z _ 0$. Then we can write

\[f(z) = (z-z_0)^kg(z)\]

where $g(z) \ne 0$ and is holomorphic in some neighbourhood of $z _ 0$. Then it follows that

\[f'(z) = k(z-z_0)^{k-1} g(z) + (z-z_0)^k g'(z)\]

so taking the quotient we have

\[h(z) = \frac{f'(z)}{f(z)} = \frac{k}{z - z_0} + \frac{g'(z)}{g(z)}\]

where $\frac{g’(z)}{g(z)}$ is holomorphic in some neighbourhood of $z _ 0$. Similarly if $f$ has a pole of order $k$ at $z _ 0$, then (abusing notation by reusing)

\[f(z) = (z-z_0)^{-k}g(z)\]

for some holomorphic function $g(z)$ in a neighbourhood of $z _ 0$. Then

\[h(z) = \frac{f'(z)}{f(z)} = \frac{-k}{z - z_0} + \frac{g'(z)}{g(z)}\]

Hence

  • If $f$ has a zero of order $k$, then $f’/f$ has simple pole with residue $k$
  • If $f$ has a pole of order $k$, then $f’/f$ has simple pole with residue $-k$

Then by the residue theorem

\[\begin{aligned} \frac{1}{2\pi i} \int_ \gamma h(z) \text dz &= \sum_ {z_ 0 \in S} \text{Res}_ {z_ 0}(f) \\\\ &= \sum_ {z_ 0 \text{ zero of }f} (\text{multiplity of zero}) - \sum_ {z_0 \text{ pole of }f} (\text{multiplity of pole}) \\\\ &= N - P \end{aligned}\]


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