## Definitions

On notation: for an analytic function \(f\) expanded as a power series about \(a\), write \(v_a(f)\) as the \(a{\hbox{-}}\)adic valuation of \(f\): expanding \(f(z) = \sum_{k\in {\mathbf{Z}}} a_k (z-a)^k\) about \(a\), define \(v_a(f) = n\) iff \(a_n\neq 0\) but \(a_{\leq n} = 0\). In other words, \(v_a\) is the lowest power of \((z-a)\) occurring in a Laurent expansion of \(f\) about \(a\).

A point \(z_0\) is an **isolated singularity** if \(f(z_0)\) is undefined but \(f(z)\) is defined in a punctured neighborhood \(D(z_0)\setminus\left\{{z_0}\right\}\) of \(z_0\).

There is a classification of isolated singularities:

- Removable singularities, so \(v_a(f) \in [0, \infty)\)
- Poles, so \(v_a(f) \in (-\infty, 0)\)
- Essential singularities, so \(v_a(f) = -\infty\), e.g. \(\sin(z^{-1})\) at \(z=0\).

Not all singularities are isolated, and thus don’t fall into this classification. One may also have **branch singularities**, e.g. \(\operatorname{Log}(z)\) at \(z=0\). \(f(z) \coloneqq z^{1\over 2}\) has a singularity at zero that does not fall under this classification – the point \(z=0\) is a **branch singularity** and \(f\) admits no Laurent expansion around \(z=0\). A similar example: \(\qty{z(z-1)}^{1\over 2}\) has two branch singularities at \(z=0, 1\).

Singularities can be classified by Laurent expansions \(f(z) = \sum_{k\in {\mathbf{Z}}} c_k z^k\):

- Essential singularity: infinitely many negative terms.
- Pole of order \(N\): truncated at \(k = -N\), so \(c_{N-\ell} = 0\) for all \(\ell\).
- Removable singularity: truncated at \(k=0\), so \(c_{\leq -1} = 0\).

Isolated singularities can also be classified by their limiting behavior near the singularity:

- \(\lim_{z\to z_0} f(z) < \infty\): removable (equivalently: bounded in a neighborhood).
- \(\lim_{z\to z_0} f(z) = \infty\): pole.
- \(\lim_{z\to z_0} f(z)\) does not exist: essential.

For any \(f\) holomorphic on an unbounded region, we say \(z=\infty\) is a singularity (of any of the above types) of \(f\) if \(g(z) \coloneqq f(1/z)\) has a corresponding singularity at \(z=0\).

## Removable

If \(z_0\) is a singularity of \(f\). then \(z_0\) is a **removable singularity** iff there exists a holomorphic function \(g\) such that \(f(z) = g(z)\) in a punctured neighborhood of \(z_0\). Equivalently,
\begin{align*}
\lim_{z\to z_0}(z-z_0) f(z) = 0
.\end{align*}
Equivalently, \(f\) is bounded on a neighborhood of \(z_0\). Equivalently, \(v_{z_0}(f) \geq 0\)

- \(f(z) \coloneqq\sin(z)/z\) has a removable singularity at \(z=0\), and one can redefine \(f(0) \coloneqq 1\).
- If \(f(z) = p(z)/q(z)\) with \(q(z_0)=0\) and \(p(z_0)=0\), then \(z_0\) is removable with \(f(z_0)\coloneqq p'(z_0)/q'(z_0)\).

Suppose \(f\) is holomorphic on \(\Omega\setminus\left\{{z_0}\right\}\). TFAE:

- \(z_0\) is a pole of order \(0\).
- \(z_0\) is a removable singularity of \(f\).
- There exists some neighborhood of \(z_0\) on which \(f\) is bounded.
- \((z-a)f(z) \overset{z\to z_0}\longrightarrow 0\)
- \(f\) admits a holomorphic extension \(F\) to all of \(\Omega\)
- \(f\) admits a continuous extension \(F\) to all of \(\Omega\).
- \(f\) admits a Laurent expansion about \(z_0\) with vanishing principal part, i.e. \(f(z) = \sum_{k\geq 0}c_k (z-z_0)^k\).

Take \(\gamma\) to be a circle centered at \(z_0\) and use \begin{align*} f(z) \coloneqq\int_\gamma { f(\xi) \over \xi - z} \,dx .\end{align*} This is valid for \(z\neq z_0\), but the right-hand side is analytic. (?)

A singularity of a holomorphic function is removable if and only if the function is bounded in some punctured neighborhood of the singular point.

## Essential

A singularity \(z_0\) is *essential* iff it is neither removable nor a pole. Equivalently, a Laurent series expansion about \(z_0\) has a principal part with infinitely many terms.

\(f(z) \coloneqq e^{1/z}\) has an essential singularity at \(z=0\), since we can expand and pick up infinitely many negative terms: \begin{align*} e^{1/z} = 1 + {1\over z} + {1\over 2! z^2} + \cdots .\end{align*} In fact there exists a neighborhood of zero such that \(f(U) = {\mathbf{C}}\setminus\left\{{0}\right\}\). Similarly \(g(z) \coloneqq\sin\qty{1\over z}\) has an essential singularity at \(z=0\), and there is a neighborhood \(V\) of zero such that \(g(V) = {\mathbf{C}}\).

The singularities of a rational function are always isolated, since there are finitely many zeros of any polynomial. The function \(F(z) \coloneqq\operatorname{Log}(z)\) has a singularity at \(z=0\) that is **not** isolated, since every neighborhood intersects the branch cut \((-\infty, 0) \times\left\{{ 0 }\right\}\), where \(F\) is not even defined. The function \(G(z) \coloneqq 1/\sin(\pi/z)\) has a non-isolated singularity at 0 and isolated singularities at \(1/n\) for all \(n\).

## Zeros

A subset of \({\mathbf{R}}^n\) is closed and bounded iff it is sequentially compact. Equivalently, every bounded sequence has a convergent subsequence.

Why this is useful: every infinite subset of a disk has a limit point. So e.g. if \(f\) is holomorphic and has infinitely many zeros in \({\mathbb{D}}\), \(f\) is identically zero by the identity principle.

A **zero** of an analytic function on a domain \(\Omega\) is any \(z_0\) such that \(f(z_0)=0\), with no further conditions. If \(f\) is analytic and not identically zero on \(\Omega\) with \(f(z_0) = 0\), then there exists a neighborhood \(U\ni z_0\) and function \(g\) that is holomorphic and nonvanishing on \(U\) such that
\begin{align*}
f(z) = (z-z_0)^n g(z)
.\end{align*}
We refer to \(z_0\) as a **zero of order \(n\)**. Equivalently, \(f^{(n-1)}(z_0)=0\) but \(f^{(n)}(z) \neq 0\), so the Laurent expansion has the form \(f(z) = \sum_{k\geq n} c_k (z-z_0)^k\) where \(c_n\neq 0\).

On terminology: if the order of \(z_0\) for \(f\) is \(n\), we say \(f\) **vanishes to order \(n\)**.

Use that \(\Omega\) is connected to find some neighborhood \(U\) on which \(f\) is not identically zero. WLOG assume \(z_0=0\). Expand \(f\) as an honest power series at \(z_0\) to write \begin{align*} f(z) = \sum_{k\geq 0}c_k z^k = z^n\qty{c_n + c_{n+1}z + \cdots} \coloneqq z^n g(z) ,\end{align*} where \(a_n\) is the minimal nonvanishing coefficient. Since \(a_n\neq 0\), we have \(\lim_{z\to z_0} g(z) = a_n \neq 0\), so \(g\) is nonvanishing in some neighborhood of \(z_0\). Uniqueness follows from writing \begin{align*} z^n g(z) = z^m h(z) \implies g(z) = z^{m-n} h(z) ,\end{align*} assuming \(m>n\), but then taking \(z\to z_0 =0\) on the RHS yields \(g(z) = 0\), a contradiction.

If \(f:{\mathbf{C}}\to {\mathbf{C}}\) is holomorphic and not identically zero, then \(f\) has isolated zeros.

Suppose not, then pick a limit point \(z_0\) with \(f(z_0)=0\) with a sequence \(\left\{{z_k}\right\}\to z_0\) where \(f(z_k) = 0\) for all \(k\). Expand \(f\) in a Laurent series; since \(f\not\equiv 0\) there is a smallest nonzero coefficient \(c_m\): \begin{align*} f(z) = \sum_{k\geq m}c_k (z-z_0)^k = c_m(z-z_0)^m \cdot\qty{1 + \sum_{k\geq 1}c_k' (z-z_0)^k } \coloneqq c_m(z-z_0)^m \cdot(1 + g(z-z_0)) .\end{align*} Note \(g(z-z_0)\overset{z\to z_0}\longrightarrow 0\), and since \(z_k\to z_0\) we can find \(k\gg 1\) such that \(g(z_k - z_0) < {\varepsilon}\). In particular, for \(k\) large enough, \(1 < 1+g(z_k - z_0) 1 + {\varepsilon}\), but this contradicts \(f(z_k) = 0\): \begin{align*} 0 = f(z_k) = c_m(z_k - z_0)^m (1 + g(z_k - z_0)) \neq 0 .\end{align*}

\(\contradiction\)

If \(f,g\) are holomorphic and \(f=g\) on any set with a limit point, then \(f\equiv g\).

The proof follows from the fact that \(f-g\) is holomorphic and has nonisolated zeros.

## Poles

Let \(f\) be a meromorphic function with an isolated singularity at \(z_0\). TFAE:

- \(z_0\) is a pole of \(f\) of order \(n\).
- \({\left\lvert {f(z)} \right\rvert}\overset{z\to z_0}\longrightarrow \infty\)
- \(z_0\) is a zero of order \(n\) of \(g(z) \coloneqq{1\over f(z)}\), and \(g\) is holomorphic in a neighborhood of \(z_0\)
- \(f(z) = (z-z_0)^{-n}h(z)\) where \(h\) is holomorphic in a punctured neighborhood of \(z_0\).
- \(f\) admits a Laurent expansion of the form \begin{align*} f(z) = \sum_{k\geq -n} c_k (z-z_0)^k, && c_{-n}\neq 0 .\end{align*}

Any pole admits a neighborhood where \(f\) is nonvanishing, and in fact bounded below.

In this case there exists a minimal \(n\) and a holomorphic \(h\) such that
\begin{align*}
f(z) = \qty{z-z_0}^{-n} h(z)
.\end{align*}
Such an \(n\) is the *order* of the pole. A pole of order 1 is said to be a *simple pole*.

Use that \(z_0\) is a zero of \(1/f\) to write \begin{align*} {1\over f(z) } = (z-z_0)^n g(z) ,\end{align*} for \(h\) holomorphic and nonvanishing in a neighborhood of \(z_0\). Taking reciprocals yields \begin{align*} f(z) = (z-z_0)^{-n} h(z) && \quad h(z) \coloneqq{1\over g(z)} .\end{align*}

Claim: if \(f\) has a pole of order \(m\) at \(z_0\), then \(g(z) \coloneqq f(z^2)\) has a pole of order \(2m\) at \(z_0\). WLOG assume \(z_0=0\). Note that this is clear by multiplying Laurent expansions about \(z_0\): \begin{align*} f(z) = \sum_{k\geq -m} c_k z^k \implies g(z) = \sum_{k\geq -m} c_k z^{2k} = {c_{-m} \over z^{2m}} + \cdots .\end{align*} Using the other characterization, write \(f(z) = z^{-m} h(z)\) with \(g\) holomorphic and nonzero in a neighborhood \(U\) of \(z_0\), so in particular \(h(0) \neq 0\). Then \(f(z^2) = z^{-2m} h(z^z)\) and \(h(z^2)\mathrel{\Big|}_{z=0} = h(0) \neq 0\).

If \(f\) has a pole of order \(n\) at \(z_0\), then there exist a holomorphic \(G\) in a neighborhood of \(z_0\) such that \begin{align*} f(z) = {a_{-n} \over (z-z_0)^n } + \cdots + {a_{-1} \over z-z_0} + G(z) \coloneqq P(z) + G(z) .\end{align*}

The term \(P(z)\) is referred to as the *principal part of \(f\) at \(z_0\)* consists of terms with negative degree, and the *residue* of \(f\) at \(z_0\) is the coefficient \(a_{-1}\).