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 $z _ 0$?
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?
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 $f$”?
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 $\text{Res} _ {z _ 0} (f)$?
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 $z _ 0$.
- In this case, there $\exists k > 0$ and a holomorphic function $g$ such that $g(z _ 0) \ne 0$ and $f(z) = (z-z _ 0)^k g(z)$.
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.