Notes - Complex Analysis MT23, Zeroes, singularities and poles


Flashcards

Suppose

  • $U$ is an open set
  • $g : U \to \mathbb C$ is holomorphic on $U$

Let $S = \{z \in U \mid g(z) = 0\}$. Can you classify the points $z _ 0 \in S$ into two cases?


  • $z _ 0$ is isolated, $g$ is non-zero on some disk about $z _ 0$ except at $z _ 0$ itself.
  • $g = 0$ in a neighbourhood of $z _ 0$

Suppose

  • $U$ is an open set
  • $g : U \to \mathbb C$ is holomorphic on $U$

Let $S = \{z \in U \mid g(z) = 0\}$. Then the points $z _ 0 \in S$ fall into two cases:

  • $z _ 0$ is isolated, $g$ is non-zero on some disk about $z _ 0$ except at $z _ 0$ itself.
  • $g = 0$ in a neighbourhood of $z _ 0$

In the former case, how can you write an expression for $g$ about $z _ 0$ in terms of holomorphic function on $U$ without the property that $f(z _ 0) = 0$?


There $\exists k \in \mathbb Z, k > 0$ and a holomorphic function $g _ 1$ such that

\[g(z) = (z-z_0)^k g_1(z)\]

and $g _ 1(z _ 0) \ne 0$.

Suppose

  • $U$ is an open set
  • $g : U \to \mathbb C$
  • $g(z _ 0) = 0$, and non-zero on punctured disc about $z _ 0$ (i.e. an isolated zero)

Can you define the multiplicity of the zero $z _ 0$ (or sometimes called “the order of vanishing”)?


Unique integer $k > 0$ such that

\[g(z) = (z-z_0)^k g_1(z)\]

where $g _ 1$ holomorphic and $g _ 1(z _ 0) \ne 0$.

Define what it means for a function $f$ to have an isolated singularity.


$f$ is holomorphic on $B(z _ 0, r) \setminus \{z _ 0\}$ for some $r > 0$ but it is not holomorphic at $z _ 0$.

Define what it means for a function $f$ with isolated singularity $z _ 0$ to be a removable singularity?


It is possible to redefine $f(z _ 0)$ so that $f$ becomes holomorphic at $z _ 0$.

Suppose

  • $f$ has isolated singularity $z _ 0$
  • $f$ is not bounded near $z _ 0$
  • $1/f$ has a removeable singularity at $z _ 0$

What is this called?


A pole of $f$ at $z _ 0$.

Define what it means for a function $f$ with isolated singularity $z _ 0$ to have a pole.


  • $f$ has isolated singularity $z _ 0$
  • $f$ is not bounded near $z _ 0$
  • $1/f$ has a removeable singularity at $z _ 0$

Suppose $f$ has an isolated singularity $z _ 0$ that is not removeable or a pole. What is this called?


An essential singularity.

Define what it means for a function $f$ with isolated singularity $z _ 0$ to have an essential singularity?


It is not a remoeable or a pole.

Suppose

  • $f$ has an isolated singularity $z _ 0$.
  • $z _ 0$ is a pole (i.e. it is a removeable singularity of $1/f$).

Can you define the order of such a pole?


The unique integer $m$ such that

\[(1/f)(z) = (z - z_0)^m g(z) \iff f(z) = (z-z_0)^{-m} \cdot 1/g(z)\]

where $g$ is a holomorphic and nonzero at $z _ 0$.

What does it mean for $f$ to have a simple pole at an isolated singularity $z _ 0$?


\[(1/f)(z) = (z-z_0)g(z)\]

for some nonzero holomorphic function $g$, defined on a neighbourhood of $z _ 0$.

Suppose $f$ has an isolated singularity at point $z _ 0$. Can you give the characterisation of $z _ 0$ being a pole in terms of limits?


\[z_0 \text{ is a pole} \iff |f(z)| \to \infty \text{ as } z \to z_0\]

Can you define what it means for a function $f : U \to \mathbb C$ to be a meromorphic function?


$f$ only has isolated singularities, all of which are poles.

What is the terminology used to describe a function $f : U \to \mathbb C$ where every singularity of $f$ is isolated and a pole?


A meromorphic function.

Suppose $f$ has a pole of order $m$ at $z _ 0$. What is the Laurent series for $f$ about $z _ 0$?


A set of coefficients $(c _ n)$ such that

\[f(z) = \sum_{n \ge -m} c_n (z - z_0)^n\]

for all $z \in B(z _ 0, r)$ where $r > 0$.

Suppose $f$ has a pole of order $m$ at $z _ 0$. Can you define the principal part of $f$ at $z _ 0$, denoted $P _ {z _ 0}(f)$, and use this to define a useful function $g$ which is sort of like “the holomorphic parts of $f$”?


\[\sum^{-1} _{n =-k} c _n (z - z _0)^n\]

Then

\[g(z) = f(z) - P\\_{z_0}(f)(z)\]

(this function is useful for its applications to the residue theorem)

Quickly prove a baby version of the residue theorem, that if

  • $f$ is a meromorphic function (all singularities are isolated singularities, and every one of these is a pole),
  • $\gamma : [0, 1] \to U$ is a closed path with $\gamma^\star \cap S = \emptyset$

Then

\[\int_\gamma f(z) \text d z = 2\pi i \sum_{z_0 \in S} \left( \text{Res}\_{z\_0}(f) \cdot I(\gamma, z\_0) \right)\]

(here $S$ denotes the set of singularities).


Consider

\[g(z) = f(z) - \sum_{z_0 \in S} P\_{z\_0}(f)(z)\]

which will be holomorphic on $U$ for reasons. Then

\[\begin{aligned} \int_\gamma f(z) \text dz &= \int_\gamma g(z) \text d z + \sum_{z_0 \in S} \int_\gamma P_{z_0}(f)(z) \text d z \\\\ &= \sum_{z_0 \in S} \int_\gamma P_{z_0}(f)(z) \text d z \\\\ &= \sum_{z_0 \in S} \text{Res}_{z_0}(f) \int_\gamma \frac{1}{z - z_0} \text dz \\\\ &= \sum_{z_0 \in S} \left(\text{Res}_{z_0}(f) \cdot 2\pi i I(\gamma, z_0)\right) \\\\ \end{aligned}\]

Suppose that $f$ has a pole of order $m$ at $z _ 0$. Can you give a formula for $\text{Res} _ {z _ 0} (f)$?


\[\lim_{z \to z_0} \frac{1}{(m-1)!} \frac{\text d^{m-1}\\,}{\text dz^{m-1}\\,} \left( (z-z_0)^m f(z) \right)\]

Proofs

Suppose:

  • $U$ is an open set
  • $f : U \to \mathbb C$ holomorphic
  • $S = \{z \in U \mid g(z) = 0\}$

Quickly prove that the points $z _ 0 \in S$ fall into two cases:

  • $z _ 0$ is isolated, $f$ is non-zero on some disk about $z _ 0$ except at $z _ 0$ itself.
    • In this case, there $\exists k > 0$ and a holomorphic function $g$ such that $g(z _ 0) \ne 0$ and $f(z) = (z-z _ 0)^k g(z)$.
  • …or, $g = 0$ in a neighbourhood of $z _ 0$.

Pick some $z _ 0 \in S$. Since $f$ is holomorphic on $U$, it is in particular analytic, so $\exists r > 0$ s.t.

\[f(z) = \sum^\infty_{k = 0} c_k(z-z_0)^k\]

$\forall z \in B(z _ 0, r)$. Then either $c _ k = 0$ for all $k$, in which case $f = 0$ on $B(z _ 0, r)$. Otherwise, define

\[k := \min\\{n \in N \mid c_n \ne 0\\}\]

Then define

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

so that $f(z) = (z-z _ 0)^k g(z)$. $g(z)$ is clearly holomorphic on $B(z _ 0, r) \setminus \{z _ 0\}$, but we can make it holomorphic on all of $B(z _ 0, r)$ by setting $g(z _ 0) = c _ k$.

Then since $g$ is continuous at $z _ 0$ and $g(z _ 0) \ne 0$, $\exists \varepsilon > 0$ such that $g(z) \ne 0$ for all $z \in B(z _ 0, \varepsilon)$.

Since $(z-z _ 0)^k$ vanishes only at $z _ 0$, it follows $f(z) = (z-z _ 0)^k g(z)$ is non-zero on $B(z _ 0, \varepsilon) \setminus \{z _ 0\}$ and hence is an isolated singularity.

Suppose $f$ has an isolated singularity at point $z _ 0$. Quickly prove that

\[z_0 \text{ is a pole} \iff |f(z)| \to \infty \text{ as } z \to z_0\]

Suppose $z _ 0$ is a pole. Then

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

for some holomorphic $g$ where $g(z _ 0) \ne 0$ and $k > 0$. Then

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

Since $g(z)$ is bounded away from $0$ near $z _ 0$, and $(z-z _ 0)^{-k} \to \infty$, $ \vert f(z) \vert \to \infty$ as $z \to z _ 0$.

Now suppose that $ \vert f(z) \vert \to \infty$ as $z \to z _ 0$. Then

\[\frac{1}{f(z)} \to 0\]

as $z \to z _ 0$. Thus $1/f$ has a removeable singularity at $z _ 0$, so $f$ has a pole at $z _ 0$ (this is the definition of a pole, but to see it explicitly consider now the fact that $1/f$ is holomorphic, so has a series expansion in $(z-z _ 0)$, then taking reciprocals gives the form).

Quickly prove that if $f$ has a pole of order $m$ at $z _ 0$, then $\text{Res} _ {z _ 0}(f)$ is given by

\[\lim_{z \to z _ 0} \frac{1}{(m-1)!} \frac{\text d^{m-1} }{\text dz^{m-1} } \left( (z-z_ 0)^m f(z) \right)\]

We have $f(z) = \sum _ {n \ge -m} c _ n (z - z _ 0)^n$ for $z$ sufficiently close to $z _ 0$. Hence

\[(z - z_0)^m f(z) = c_{-m} + c_{-m + 1} (z - z_0) + \cdots + c_{-1} (z - z_0)^{m-1} + \cdots\]

and then differentiating $m-1$ times gives us:

\[(m-1)! \cdot c_{-1} + m! \cdot c_0 (z-z_0) + (m+1)! \cdot c_1 (z-z_0)^2\]

Hence taking the limit as $z \to z _ 0$ and dividing through by $(m-1)!$ gives us

\[c_{-1 } = \lim_ {z \to z_ 0} \frac{1}{(m-1)!} \frac{\text d^{m-1} }{\text dz^{m-1} } \left( (z-z_ 0)^m f(z) \right)\]

as required.




Related posts