Meromorphic Function

We talk about singularities at the complex plane.

• Removable singularities.

• Poles.

• Essential singularities.

Zeros and poles

We start from zeros and poles.

Actually, by the Theorem for zero accumulation, if we have an accumalation point of a sequence of zeros, then a holomorphic fucntion \(f\) is identically zero on region \(\Omega\) (here must be connected). So for a non-trivial holomorphic function on \(\Omega\), its zeros are isolated, i.e. have no limit point. To be elaborate, if \(f(z_0)=0\), then \(\exists \delta>0\), such that \(f(z)\neq 0,\forall z\in D_\delta(z_0)-\{z_0\}\).

Based on the above analysis, we have the following configuration for zeros with power series.

Theorem for Zeros with power series

Assume \(f\) is holomorphic on a region \(\Omega\), \(f(z_0)=0\). For a neighborhood \(U\subset \Omega\) of \(z_0\), \(f\) does not vanish identically on \(U-\{z_0\}\). Then there exists a unique positive interger \(n\) such that

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

where \(g(z)\) is a holomorphic function which does not vanish on \(U\). We call \(f\) has zero of order \(n\) (multiplicity \(n\)) at \(z_0\).

Proof

• Existence.

\(f\) is holomorphic on \(U\), so we have power series at \(z_9\)

\[ f(z)=\sum_{m=0}^\infty a_m (z-z_0)^m. \]

Since \(f\) is non-trivial, there exsits a smallest number \(n\) such that \(a_n\neq 0\), otherwise \(f\) is identically zero. So

\[ f(z)=(z-z_0)^n\sum_{m=n}^\infty a_m (z-z_0)^{m-n}=(z-z_0)^n g(z) \]

• Uniqueness. Assume \(\exists n,k\) such that \(n<k\) and

\[ \begin{align*} f(z)=(z-z_0)^n\sum_{m=n}^\infty a_m (z-z_0)^{m-n}&=(z-z_0)^k\sum_{m=k}^\infty a_m (z-z_0)^{m-k}\\ \Rightarrow \quad \sum_{m=n}^\infty a_m (z-z_0)^{m-n} &= (z-z_0)^{k-n}\sum_{m=k}^\infty a_m (z-z_0)^{m-k} \end{align*} \]

then substitute in \(z_0\), we have the right-hand expression equal \(0\) while the left-hand does not.

Define a function \(f\) defined on a deleted neighborhood of \(z_0\) is holomorphic and has a pole at \(z_0\), if \(1/f\) has a zero at \(z_0\) and holomorphic on all the neighborhood of \(z_0\).

Theorem for Poles with power series

Assume \(f\) is holomorphic on a deleted neighborhood of \(z_0\), and has a pole at \(z_0\). Then there exsits a unique positive interger \(n\) such that

\[ f(z)=(z-z_0)^{-n} h(z) \]

where \(h(z)\) is a non-vanishing holomorphic function on the full neighborhood of \(z_0\). We call \(f\) has poles of order \(n\) (multiplicity \(n\)) at \(z_0\). Moreover

\[ \begin{align} f(z)=\sum_{k=1}^n \frac{a_{-k}}{(z-z_0)^k} + G(z)\label{configuration of poles} \end{align} \]

where \(G(z)\) is still a holomorphic function on the full neighborhood of \(z_0\), but might vanish at \(z_0\). Here \(\sum\limits_{k=1}^n \frac{a_{-k}}{(z-z_0)^k}\) is called the principle part of \(f\) at the pole \(z_0\).

Proof

Just use Theorem for Zeros with power series, note that

\[ h(z)=\frac{1}{g(z)}\neq 0,\quad \forall z\in U. \]

Since \(h(z)\) is holomorphic, we apply power series expansion again to \(h(z)\) and have the result needed.

Naturally we want to apply the closed loop integral on the function with poles, which by the above configuration equation \(\ref{configuration of poles}\), only

\[ \frac{a_{-1}}{(z-z_0)} \]

would yields non-zero value, since other parts have a primitive on \(U-\{z_0\}\). Define residue to be the coefficient of this item, i.e. \(\text{res}_{z_0} f=a_{-1}\). Then actually for a closed loop \(\gamma\) in \(U-\{z_0\}\),

\[ \int_\gamma f(z)dz=2\pi i \text{ res}_{z_0}f. \]

To elaborate, we have the following theorem.

Residue Formula

Theorem of Residue formula

Assume \(f\) is holomorphic on a open set containing a circle \(C\) and its interior, except for a pole at \(z_0\). Then

\[ \int_C f(z)dz=2\pi i\text{ res}_{z_0}f. \]

The above theorem could be extended to finitely many poles in a open set.

A lot of questions occur on acquaring the residue of functions with poles.

Calculation of residue (a-1)

Calculation of Residue

For simple pole, we have

\[ \text{res}_{z_0} f=\lim_{z\rightarrow z_0}(z-z_0)f(z), \]

For pole of multiple order \(n\) (\(n\geq 1\)), we have

\[ \text{res}_{z_0} f= \lim_{z\rightarrow z_0} \frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}}(z-z_0)^nf(z). \]

Removable Singularies

Definition of removable singularities

Assume \(f\) is holomorphic on open set \(\Omega\) except for \(z_0\). If we could define \(f\) at \(z_0\) such that \(f\) is holomorphic on \(\Omega\), then we call \(z_0\) a removable singularity for \(f\).

How to define? We could use inspirations from Sequence of holomorphic functions or in terms of integrals, that is, create a function that agree with original function on \(\Omega-\{z_0\}\) and itself has a value at \(z_0\). This is achieved by Cauchy's Integral formula.

Riemann's theorem on removable singularities

Assume \(f\) is holomorphic on open set \(\Omega\) except for \(z_0\). If \(f(z)\) is bounded on \(\Omega-\{z_0\}\), then \(z_0\) is a removable point.

ProofWrong version

Choose \(R\) and we have a circle \(C_R(z_0)\subset \Omega\), and an open disc \(D_R(z_0)\), define

\[ F(z)=\int_{C_R(z_0)}\frac{f(\zeta)}{(\zeta-z)}dz ,\quad z\in D_R(z_0), \]

which has a definition, for each \(\zeta\in C_R(z_0)\), the integrated function \(\frac{f(\zeta)}{(\zeta-z)}\) is holomorphic for all \(z\in D_R(z_0)\), so by function in terms of integrals, \(F(z)\) is holomorphic for all \(z\in D_R(z_0)\). Then if we find out

\[ \begin{align} F(z)=f(z), \quad \forall z\in D_R(z_0)\label{analytic-part} \end{align} \]

then \(F\) is a analytic continuation of \(f\), and could define \(f(z_0)=F(z_0)\).

• Show that equation \(\ref{analytic-part}\) holds. That is,

\[ f(z)=\int_{C_R(z_0)}\frac{f(\zeta)}{(\zeta-z)}dz, \]

which is similar to Cauchy's integral formula. We have to formulate a toy contour and actually

\[ \int_{C_R(z_0)}\frac{f(\zeta)}{(\zeta-z)}dz=\int_{C_\varepsilon(z)}\frac{f(\zeta)}{(\zeta-z)}dz+\int_{C_\varepsilon(z_0)}\frac{f(\zeta)}{(\zeta-z)}dz. \]

The first item is \(2\pi i f(z)\) and the second item equals \(0\) since \(f\) is bounded

\[ \left|\int_{C_\varepsilon(z_0)}\frac{f(\zeta)}{(\zeta-z)}dz\right| \leq \frac{\left|\sup\limits_{z\in \Omega-\{z_0\} }f(z)\right|}{|z-z_0|/2}\cdot 2\pi \varepsilon\rightarrow 0. \]

and we are done.

We could also define

\[ g(z)=\begin{cases} (z-z_0)f(z),\quad &z\in \Omega-\{z_0\}\\ 0,\quad &z=z_0 \end{cases}. \]

So \(g(z)\) is holomorphic on \(\Omega\) because at \(z_0\), we have a wrong version:

\[ \lim_{z\rightarrow z_0}\frac{g(z)-g(z_0)}{z-z_0}=\lim_{z\rightarrow z_0}f(z) \]

cause we have not define the limit of \(f(z)\) at \(z_0\). Actually, we only have

\[ |g(z)|\leq M(z-z_0)\rightarrow 0,\quad (z\rightarrow z_0) \]

So \(\lim_{z\rightarrow z_0}g(z)=0\).

However, if we already have the theorem, then we could use the above estimate to show that \(g(z)\) is bounded, thus holomorphic.