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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 \(z_0\).

  • 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 \(z_0\).

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.

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.

Corollary: Characteristics of poles

Assume \(f\) is holomorphic on a region except for \(z_0\). Then \(z_0\) is a pole iff \(\lim\limits_{z\rightarrow z_0}|f(z)|=\infty\).

  • Necessary.

If \(z_0\) is a pole, then \(1/f\) has a zero at \(z_0\), i.e. \(\lim\limits_{z\rightarrow z_0}1/f(z)=0\), i.e. \(\lim\limits_{z\rightarrow z_0}|f(z)|=\infty\).

  • Sufficient.

If \(\lim\limits_{z\rightarrow z_0}|f(z)|=\infty\), then \(\frac{1}{f}\) is bounded near \(z_0\), so it has a removable singularity there. So define \(1/f(z)|_{z=z_0}=\lim\limits_{z\rightarrow z_0}1/|f(z)|=0\). So it has a pole at \(z_0\).

Essential Singularities

Casorati-Weierstrass for essential singularities

Suppose \(f\) is holomorphic in a punctured disc \(D_r(z_0)-\{z_0\}\) and has an essential singularity at \(z_0\). Then, the image of \(D_r(z_0)-\{z_0\}\) under \(f\) is dense in \(\mathbb{C}\).

We argue by contradiction.

Assume the image of \(D_r(z_0)-\{z_0\}\) is not dense in \(\mathbb{C}\), then \(\exists w\in \mathbb{C}\), \(\exists \delta>0\), such that \(\forall z\in D_r(z_0)-\{z_0\}\), \(|w-f(z)|>\delta\). Define

\[ g(z)=\frac{1}{f(z)-w}. \]

which is holomorphic in \(D_r(z_0)-\{z_0\}\), and bounded by \(\frac{1}{\delta}\), so it has a removable singularity at \(z_0\). Now we consider the exact value of \(g(z_0)\).

If \(g(z_0)=0\), it is a pole for \(f(z)-w\) at \(z_0\), contradicts! If \(g(z_0)\neq 0\), then \(f(z_0)=w+1/g(z_0)\), and

\[ f(z)=w+1/g(z), \quad z\in D_r(z) \]

so \(f(z)\) is holomorphic at \(z_0\), which also contradicts!

Functions with only poles

Definition for meromorphic function

A function \(f\) on an open set \(\Omega\) is meromorphic, if there exsits a sequence of points \(\{z_n\}\) that has no limit points in \(\Omega\) and such that

(1) \(f\) is holomorphic on \(\Omega-\{z_n\}\)

(2) \(f\) has poles at points \(\{z_n\}\).

Notice that the poles are isolated by analytic continuation, otherwise \(f\equiv 0\). If \(\Omega\) is bounded or compact, then \(z_n\) must be finite number. So if \(z_n\) has infinite many number, we must have \(\Omega\) extend to \(\infty\). Here comes the discussion about singularites at infinity.

singularities at infinity

If \(f\) is holomorphic for all large value of \(z\), define \(F(z)=f(\frac{1}{z})\), which is holomorphic in a deleted neighborhood of the origin. We say that \(f\)

(i) has a pole at \(\infty\) if \(F\) has a pole at \(0\).

(ii) has an essential singularity at \(\infty\) if \(F\) has an essential singularity at \(0\).

(iii) has a removable singularity at \(\infty\) if \(F\) has a removable singularity at \(0\).

A meromorphic function in the complex plane that is either holomorphic at infinity or has a pole at infinity is said to be meromorphic in the extended complex plane.

Now we have a description for meromorphic function.

Description of meromorphic function

The meromorphic functions in the extended complex plane are the rational functions.

  • For singularities at infinity. Let \(F(z)=f(\frac{1}{z})\), so it could also be expressed by \(F(z)=F_\infty(z)+G_\infty(z)\) where \(F_\infty(z)\) is the principle part and \(G_\infty(z)\) is holomorphic at \(0\) (at all other points as well). Note here their forms are

  • For singularities less than infinity, we have only finite number of poles, denoted by \(\{z_i\}_{1\leq i\leq N}\). This is because \(f(\frac{1}{z})\) has either a pole or removable singularities there, so \(\exists \delta=\frac{1}{R}\), such that \(f(1/z)\) has no other singularities in

\[ D_{\delta}(0)=\{z: |z|<\delta\}=\{u: |\frac{1}{u}|<\delta\}=\{u: |{u}|>R\}. \]

\(f(z)\) has no other singularities in \(\{u: |{u}|>R\}\), i.e has other singularities in \(D_R(0)\), which is a compact set, meaning singularities in it must have finite number. We denote its principle part \(f_i(z)\), so we have \(f(z)=f_i(z)+g_i(z)\) for each \(i\), where \(g_i\) is holomorphic at \(z_i\) (at all other points as well).

\[ F_\infty(z)=\sum_{k=-n_\infty}^{-1} a_{n_\infty}z^k,\quad G_\infty(z)=\sum_{k=0}^\infty a_\infty z^k \]

  • Define a function \(h(z)=f(z)-\sum_{i=1}^N f_i(z)-F_\infty(\frac{1}{z})\), which is an entire function.

  • Prove it is bounded. Since \(h(1/z)\) has a removable singularity near \(z=0\), thus is bounded. So \(h\) is bounded. By Liouville's theorem, \(h=c\), so

\[ f(z)=\sum_{i=1}^N f_i(z)+F_\infty(\frac{1}{z})+c \]

which is a rational function.

Example. \(f(z)=\frac{1}{\sin \pi z}\) has singularities at \(z_n=n\), which is of denumerable number. But \(f(\frac{1}{z})=\frac{1}{\sin \pi \frac{1}{z}}\) has different phenomenon at \(z=0\). For \(z=x\in \mathbb{R}\), \(f(\frac{1}{x})\rightarrow \infty\) as \(x\rightarrow 0\). But \(z=iy\), \(f(\frac{1}{iy})\rightarrow 0\) as \(y\rightarrow 0\). This is because \(f(1/z)\) has singularities at \(z_n=\frac{1}{n}\) which has a limit point at \(z=0\) (at infinity for \(f(z)\)).

Argument Principle & Applications

The basic observations are as follows. Note we have the conclusion because of the integral of a holomorphic function over a closed curve vanishes, which demonstrates the infomation of the zeros and poles inside the curve.

A simple fact: Additivity of Logarithm derivatives

Assume \(f_1,\cdots,f_N\) are holomorphic functions, then

\[ \frac{(f_1f_2)'}{f_1f_2}=\frac{f_1'f_2+f_1f_2'}{f_1f_2}=\frac{f_1'}{f_1}+\frac{f_2'}{f_2} \]

and by induction

\[ \frac{(\prod_{k=1}^N f_k)'}{\prod_{k=1}^N f_k}=\sum_{k=1}^N \frac{f_k'}{f_k}. \]

Augument Principle

Suppose \(f\) is holomorphic in an open set containing a circle \(C\) and its interior. If \(f\) has no poles and never vanishes on \(C\) (to let the integral well-defined), denote \(Z\) and \(P\) to be the number of zeros and poles inside \(C\), then

\[ \frac{1}{2\pi i}\int_C \frac{f'(z)}{f(z)}dz=Z-P. \]

We apply the above fact into an example, a holomorphic function \(f\) with a zero of order \(n\) at \(z_0\), i.e. \(f(z)=(z-z_0)^n g(z)\), where \(g(z)\) is holomorphic and does not vanish at \(z_0\). So

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

where \(G(z)\) is still holomorphic (maybe vanish at \(z_0\)). Similar method applies to a function with a pole of order \(z\) at \(z_0\), i.e. \(f(z)=(z-z_0)^{-n} h(z)\) and we have

\[ \frac{f'}{f}=\frac{-n}{z-z_0}+H(z) \]

So silimar as we discussed in integral around a closed curve, what we have is the \(\frac{\pm n}{z-z_0}\).

\(\square\)

Rouché's Theorem

As a consequence, we have the following three theorem shown by Argument Principle.

Rouché's Theorem: perturbance do not affect the number of zeros

Assume \(f\) and \(g\) are holomorphic functions on an open set containing a circle \(C\) and its interior. If

\[ |f(z)|>|g(z)|,\quad \forall z\in C, \]

then \(f\) and \(f+g\) has the same number of zeros inside the circle \(C\).

The core is to formulate a continuous function of number of zeros.

Define

\[ n_t=\frac{1}{2\pi i}\int_C \frac{f_t'(z)}{f_t(z)}dz,\quad t\in [0,1] \]

where \(f_t(z)=f(z)+tg(z)\). For \(t=0\), \(n_0\) equala the number of zeros of \(f\) inside \(C\), and for \(t=1\), \(n_1\) equala the number of zeros of \(f+g\) inside \(C\). We want to show that it is continuous. Since it is a integer, it must be constant.

In fact \(f_t\) is jointly continuous for \(t\in [0,1]\) and \(z\in C\). The only consideration is that \(f_t(z)\) does not vanish on \(C\) for all \(t\). This is apparent, because

\[ |f_t(z)|\geq |f(z)|-|tg(z)|\geq |f(z)|-|g(z)|>0. \]

and we are done.

\(\square\)

Open Mapping Theorem

We could intepret a holomorphic function \(f\) as an mapping from \(\mathbb{C}\) to \(\mathbb{C}\). A mapping is said to be open if it maps an open set to an open set. So when does a holomorphic function be an open mapping?

The following theorem interprets a mapping as a function with zeros.

Open Mapping Theorem

If \(f\) is holomorphic and non-constant in a region \(\Omega\), then \(f\) is open.

Assume \(w_0\in f(\Omega)\), and \(\exists z_0\) such that \(f(z_0)=w_0\). We need to prove that \(\exists \varepsilon\), \(B_\varepsilon(w_0)\subset f(\Omega)\). Choose \(w\in B_\varepsilon(w_0)\), we must show that \(\exists z\) such that \(f(z)=w\).

Define \(g(z)=f(z)-w\) and write \(g(z)=f(z)-w_0+(w_0-w)=F(z)+G(z)\). If \(|F(z)|>|G(z)|\) on a circle \(\{z:|z-z_0|=\delta\}\), then \(g(z)\) has the same number of zeros as \(F(z)\). Since \(f(z)=w\) has a zero, then \(f(z)=w_0\) must have a zero, so \(\exists z\) such that \(f(z)=w_0\).

What we need is to find the appropriate \(\delta\). First choose \(\delta>0\), such that \(C_\delta(z_0)=\{z:|z-z_0|=\delta\}\subset \Omega\) and \(f(z)\neq w_0\) on the circle (because zeros are isolated by non-constant condition). So this little distinction gives us to find \(\varepsilon>0\), such that \(|f(z)-w_0|\geq\varepsilon\) on the circle \(C_\delta(z_0)\). So if \(|w-w_0|<\varepsilon\), we have \(|F(z)|>|G(z)|\). By Rouché's Theorem, we are done.

\(\square\)

Maximum modulus principle

This is about the range of function on a region. A holomorphic function cannot attain its maximum in the interior of a region. Actually it is a natural sequence of open mapping theorem.

Maximum modulus principle

If \(f\) is a non-constant holomorphic function in a region \(\Omega\), then \(f\) cannot obtain a maximum in \(\Omega\).

We argue by contradiction. If \(f\) attain its maximum at \(z_0\in \Omega\), then since \(f\) is a open mapping, for \(D_\delta(z_0)\subset \Omega\), we have \(f(D_\delta(z_0))\subset f(\Omega)\) and it is an open set. So \(\exists \varepsilon>0\), \(D_{\varepsilon}(f(z_0))\subset f(D_\delta(z_0))\), that is, \(\exists f(z)\in D_{\varepsilon}(f(z_0))\)

\[ |f(z)-f(z_0)|>\varepsilon.\quad \Rightarrow |f(z)|>|f(z_0)|. \]

\(\square\)

Corollary

Assume \(f\) is holomorphic in a region \(\Omega\) with a compect closure \(\overline{\Omega}\), then

\[ \sup_{z\in \Omega}|f(z)|\leq \sup_{z\in \overline{\Omega}-\Omega} |f(z)|. \]