246A: Complex Analysis, Notes 2 – Meromorphic Functions and Properties of Analytic Functions

Complex Variables (math.CV), Math 246, Mathematics, Rowan Killip, UCLA Graduate Mathematics

246A: Complex Analysis, Notes 2 – Meromorphic Functions and Properties of Analytic Functions

In Mathematics on February 10, 2011 at 2:24 am

We continue with a discussion about meromorphic functions and the properties of analytic functions. Later notes will consider the Riemann mapping theorem, harmonic functions and the Dirichlet problem among other topics.

Definition 2.1 A function ${f}$ on an open set ${\Omega}$ is meromorphic if there exists a discrete set of points ${S = \left\{z: z \in \Omega\right\}}$ such that ${f}$ is holomorphic on ${\Omega\setminus S}$ and has poles at each ${z \in S}$ . Furthermore, ${f}$ is meromorphic in the extended complex plane if ${F(z) = f(1/z)}$ is either meromorphic or holomorphic at ${0}$ . In this case we say that ${f}$ has a pole or is holomorphic at infinity.

By collecting results from the previous section, we are immediately led to the following proposition regarding the Laurent expansions of complex valued functions.

Proposition 2.2 Let ${S}$ be the discrete set of singularities of a complex function ${f:\Omega \rightarrow \mathbb{C}}$ where ${\Omega}$ is an open set in ${\mathbb{C}}$ . For a fixed ${z_{0} \in S}$ , suppose the Laurent expansion for ${f}$ in an annulus about ${z_{0}}$ is given by ${\sum_{-\infty}^{\infty}a_{n}(z-z_{0})^{n}}$ . Then,

The function ${f}$ has a removable singularity at ${z_{0}}$ if and only if ${a_{n} = 0}$ for all ${n < 0}$ .
The function ${f}$ has a pole at ${z_{0}}$ if and only if there exists ${N \in \mathbb{Z}}$ with ${N < 0}$ such that ${a_{n} = 0}$ for all ${n < N}$ ; that is, the Laurent expansion of ${f}$ about ${z_{0}}$ has only finitely many negative terms.
The function ${f}$ has an essential singularity at ${z_{0}}$ if and only if the Laurent expansion of ${f}$ about ${z_{0}}$ has infinitely many negative terms.
Furthermore, ${f}$ is meromorphic on the extended complex plane if and only if there exists ${N \in \mathbb{Z}^{+}}$ such that ${a_{n} = 0}$ for ${n > N}$ .

Proof: The first three results follow immediately from the previous section. For the final result, consider the function ${F(z) = f(1/z)}$ . Suppose ${f}$ has an essential singularity at infinity. Then, ${F}$ has an essential singularity at the origin and item 3 and must have infinitely many negative terms. Substituting ${z}$ for ${1/z}$ , it follows that ${f}$ must have infinitely many positive terms. The other direction of this statement is similar, and our conclusion follows. ${\square}$ .

These results lead to an impressive description of meromorphic in the extended complex plane.

Theorem 2.3 Suppose ${f: \overline{\mathbb{C}} \rightarrow \overline{\mathbb{C}}}$ is a meromorphic function in the extended complex plane. Then ${f}$ is a rational function.

Proof: Let ${\left\{z_{n}\right\} \in \mathbb{C}}$ be the set of poles of the function ${f}$ . The function ${F(z) = f(1/z)}$ must be analytic in a deleted neighborhood of the origin, hence ${f}$ is analytic in a deleted neighborhood of ${\infty}$ ; the remaining region of the complex plane can contain only finitely many singularities, therefore ${\left\{z_{n}\right\}}$ must be finite. Let ${h_{k}(\frac{1}{z-z_{k}})}$ denote the principal part of the Laurent expansion of ${f}$ at the pole ${z_{k}}$ . In addition, let ${h_{\infty}}$ denote the principal part of the Laurent expansion of ${f}$ at the point at infinity. Our study of the singularities of complex functions and the Laurent expansion demonstrates that for all ${k = 1, \dots, N}$ , the principal parts ${h_{k}}$ are polynomials of finite length in ${\frac{1}{z-z_{k}}}$ ; the same holds for ${h_{\infty}}$ . Furthermore, we have

$\displaystyle g(z) = f(z) - \sum_{n=1}^{N}h(\frac{1}{z-z_{n}}) - h_{\infty}$

is entire and bounded, hence constant. It follows that ${f(z) = g(z) + \sum h(\frac{1}{z-z_{n}}) + h_{\infty}}$ is a rational function, as desired. ${\square}$

A critical result in the theory, enabled in part by the Laurent expansion allows one to calculate complex line integrals with relative ease. First, we note that there is little in the development of Cauchy’s integral formula that seems to imply that a holomorphic function is defined by its integral about a circle. In addition, Goursat’s theorem can easily be extended to rectangles and more complex polygons as well. These results seem to imply that we can extend Cauchy’s theorem and therefore much of our development of the theory to arbitrary rectifiable closed curves. Doing so will enable us to study more challenging questions with a greater number of tools at our disposal; for example, the desired extension will allow us to analytically continue Riemann’s zeta function to the entire complex plane from the small set on which it is originally holomorphic. We begin with a description of the region enclosed by a rectifiable closed curve.

Definition 2.4 Given a rectifiable closed curve ${\gamma:\left[0,1\right] \rightarrow \mathbb{C}}$ and a point ${z_{0}\not\in\gamma}$ the winding number of ${\gamma}$ about ${z_{0}}$ is given by the formula

$\displaystyle W_{\gamma}(z_{0}) = \frac{1}{2 \pi i}\oint_{\gamma}\frac{1}{z-z_{0}}dz.$

Theorem 2.5 Under the assumptions above, the following statements regarding the winding number hold:

${W_{\gamma}(z) \in \mathbb{Z}}$
The map ${z\mapsto W_{\gamma}(z)}$ is constant inside the connected component of ${\mathbb{C}\setminus\gamma}$ containing ${z}$ .
${W_{\gamma}(z) \rightarrow 0}$ as ${\left|z\right| \rightarrow \infty}$ . In other words, ${W_{\gamma}(z) \equiv 0}$ is the unbounded, connected component of ${\mathbb{C}\setminus\gamma}$ .

Before we proceed with a proof of the theorem, we note that the definition of the winding number resembles the complex logarithm. Indeed, the properties of the complex logarithm motivate the definition. Without excising a branch of the complex plane, the complex logarithm is not a well defined function. Suppose ${z\in\mathbb{C}}$ and write ${z=re^{i\theta} = re^{i(\theta + 2\pi k)}}$ for some ${\theta\in\left(-\pi, \pi\right]}$ and ${k = 1, 2, \dots}$ . Then,

$\displaystyle \begin{array}{rcl} \log(z) &=&\log(re^{i\theta}) = \log \left|r\right| + i {\rm Arg}(z) = \log \left|r\right| + i \theta \\ &=& \log(re^{i [\theta + 2 \pi k]}) = \log \left|r\right| + i {\rm Arg}(z) = \log \left|r\right| + i(\theta+2\pi k). \end{array}$

We can produce a holomorphic, well defined complex logarithm on ${\mathbb{C}}$ by defining a branch cut ${\Omega}$ ; that is, by removing a set (generally an appropriate path) ${\Omega}$ from ${\mathbb{C}}$ and restricting ${\log (z)}$ to ${\mathbb{C}\setminus\Omega}$ . However, without doing so, the value of the logarithm differs by an integral factor of ${2\pi i}$ . If we begin at the point ${1 \in \mathbb{C}}$ , traverse the unit circle, and return to ${1}$ , the complex logarithm at ${1}$ will differ by ${2 \pi i}$ . Traversing the unit circle again, we find that the complex logarithm now differs by ${2(2\pi i)}$ . It is with this in mind that we have proposed the above definition and prove the theorem. Proof:

To simplify our argument, suppose ${\gamma}$ is smooth. If it not, we subdivide ${\left[0,1\right]}$ into intervals disjoint invervals ${\left[a_{i}, a_{i+1}\right]}$ such that ${\gamma}$ is left and right differntiable at ${a_{i}}$ . Our argument then proceeds with slight modification. Consider ${W_{\gamma}(z_{0})}$ and let
$\displaystyle F(t) = \int_{0}^{t}\frac{\gamma^{\prime}(t)}{\gamma(t) - z_{0}}dt$

Then, ${F}$ is continuous on ${\left[0,1\right]}$ and differentiable with ${F^{\prime}(t) = \frac{\gamma^{\prime}(t)}{\gamma(t) - z_{0}}}$ . We hope to show that ${F(1)-F(0)=2 \pi i k}$ for some ${k\in\mathbb{Z}^{+}}$ . To this end, we write
$\displaystyle \begin{array}{rcl} \frac{d}{dt}e^{-F(t)}(\gamma(t)-z_{0}) &=& e^{-F(t)}\gamma^{\prime}(t)- F^{\prime}(t)e^{-F(t)}(\gamma(t)-z_{0})\\ &=& e^{-F(t)}\gamma^{\prime}(t) - \frac{\gamma^{\prime}(t)}{\gamma(t) - z_{0}}e^{-F(t)}(\gamma(t)-z_{0})\\ &=& 0. \end{array}$

Therefore, ${\exists C > 0}$ such that ${e^{-F(t)}(\gamma(t) - z_{0}) = C}$ . In other words, ${\gamma(t)-z_{0} = Ce^{F(t)}}$ . Since ${\gamma}$ is closed, ${\gamma(1) = \gamma(0)}$ , hence it follows that
$\displaystyle \begin{array}{rcl} Ce^{F(1)} = \gamma(1) - z_{0} = \gamma(0) - z_{0} = Ce^{F(0)}. \end{array}$

Finally, we have ${e^{F(1)} = e^{F(0)}}$ which implies ${F(1) = F(0) + 2 \pi k}$ ; that ${F(0) = 0}$ concludes the proof of our theorem.
We first show that ${z \mapsto W_{\gamma}(z)}$ is continuous inside the connection component of ${\mathbb{C}\setminus\gamma}$ . This, paired with the fact that ${W_{\gamma}(z)}$ is integer valued on the connected component proves that it must be constant there. Consider ${z_{o}, z_{1}}$ both inside the connected component of ${\mathbb{C}\setminus\gamma}$ . We have,
$\displaystyle \begin{array}{rcl} \left|W_{\gamma}(z_{0}) -W_{\gamma}(z_{1})\right| &\leq& \frac{1}{2\pi}\left|\oint_{\gamma}\frac{1}{z-z_{0}} - \frac{1}{z-z_{1}}dz\right|\\ &\leq& \frac{1}{2 \pi}\oint_{\gamma}\left|\frac{z_{0}-z_{1}}{\left(z-z_{0}\right)\left(z-z_{1}\right)}\right|dz. \end{array}$

Since ${\gamma}$ stays away from both ${z_{0}, z_{1}}$ , find that the denominator of the integrand is bounded, hence
$\displaystyle \begin{array}{rcl} \left|W_{\gamma}(z_{0}) - W_{\gamma}(z_{1})\right| &\leq& \frac{1}{2 \pi}\oint_{\gamma}\frac{\left|z_{0}-z_{1}\right|}{\left|\left(z-z_{0}\right)\left(z-z_{1}\right)\right|} dz \\ &\leq& \frac{1}{2 \pi}\oint_{\gamma}\frac{\left|z_{0}-z_{1}\right|}{\epsilon_{1}\epsilon_{2}}dz\\ &\leq& C\left|z_{0}-z_{1}\right|L(\gamma), \end{array}$

where ${C>0}$ and ${L(\gamma)}$ is the length of the curve ${\gamma}$ . Our assertion follows immediately.
Suppose ${z}$ is an element of the unbounded connected component of ${\mathbb{C}\setminus\gamma}$ and that ${\left|z\right|}$ is large. Then
$\displaystyle W{\gamma}(z) = \frac{1}{2 \pi i}\oint_{\gamma}\frac{1}{z-z_{0}} dz$

must tend to zero since ${\left|z-z_{0}\right|}$ becomes arbitrarily large. Again, the assertion is immediate. ${\square}$

We are now able to prove a general version of the celebrated residue theorem. Our derivation of the residue formula and proof of the theorem will be based on the equivalence of a collection of four statements. After exhibiting these equivalences, we will prove the general Cauchy integral theorem for simply connected sets, which will imply the first of these statements. As a result, the residue theorem will hold on simply connected sets, and we will use it to state and prove a variety of properties of analytic functions.

Definition 2.6 Suppose ${f}$ is a meromorphic function on the set ${\Omega \subset \mathbb{C}}$ with poles at ${\left\{z_{n}\right\} \in \Omega}$ . Let

$\displaystyle f(z) = \sum_{k = -m}^{\infty}a_{k}(z-z_{n})^{k}.$

The coefficient ${a_{-1}}$ is called the residue of ${f}$ at ${z_{n}}$ and is denoted ${{\rm Res}(f, z_{n}).}$

Theorem 2.7 Suppose ${f}$ has a simple pole, (i.e. a pole of order ${1}$ ) at ${z_{0}}$ . Then

$\displaystyle {\rm Res}(f, z_{0}) = \lim_{z \rightarrow z_{0}}(z-z_{0})f(z).$

Instead, if ${f}$ has a pole of order ${n > 1}$ at ${z_{0}}$ , then,
$\displaystyle {\rm Res}(f,z_{0}) = \lim_{z \rightarrow z_{0}}\frac{1}{\left(n-1\right)!}\left(\frac{d}{dz}\right)^{n-1}\left(z-z_{0}\right)^{n}f(z).$

Proof: Both statements follow directly from the Laurent expansion of ${f}$ at ${z_{0}}$ . ${\square}$

Theorem 2.8 The residue theorem Suppose ${\gamma:\left[0,1\right] \rightarrow \Omega}$ is a rectifiable, closed curve in an open set ${\Omega \subset \mathbb{C}}$ , ${f}$ is holomorphic on ${\Omega}$ and ${g}$ is meromorphic on ${\Omega}$ . Then, the following statements are equivalent:

${\oint_{\gamma}f dz = 0,}$
${f(z_{0})W_{\gamma}(z_{0}) = \frac{1}{2 \pi i}\oint_{\gamma}\frac{f(z)}{z-z_{0}} dz,}$
${\tilde{\Omega} = \left\{z \in \mathbb{C}: W_{\gamma}(z) \neq 0 \right\} \subset \Omega,}$
${\frac{1}{2 \pi i}\oint_{\gamma}g(z)dz = \sum_{w \in \Omega}{\rm Res}(g, w)W_{\gamma}(w).}$

Proof: First, we prove that statement ${1}$ implies statement ${2.}$ Let

$\displaystyle F(z, w) = \frac{f(z) - f(w)}{z-w}.$

Then, we find that

$\displaystyle \begin{array}{rcl} \frac{1}{2 \pi i}\oint_{\gamma}F(z,w) dz &=& \frac{1}{2 \pi i}\oint_{\gamma}\frac{f(z)}{z-w} dz - \frac{1}{2 \pi i}\oint_{\gamma}\frac{f(w)}{z-w} dz \\ &=& \frac{1}{2 \pi i}\oint_{\gamma}\frac{f(z)}{z-w}dz - \frac{f(w)}{2 \pi i}\oint_{\gamma}\frac{1}{z-w}dz \\ &=& \frac{1}{2 \pi i}\oint_{\gamma}\frac{f(z)}{z-w}dz - f(w)W_{\gamma}(w). \end{array}$

Since ${f}$ is holomorphic and its derivative is bounded, ${F}$ must be holomorphic in ${z}$ , hence by statement ${1}$ , ${\oint_{\gamma}F(z,w) dz = 0}$ .

On the other hand, let ${F(z) = f(z)(z-z_{0})}$ . Clearly ${F}$ is holomorphic, and statement ${2}$ then implies

$\displaystyle \begin{array}{rcl} F(z_{0})W_{\gamma}(z_{0}) &=& \frac{1}{2 \pi i}\oint_{\gamma}\frac{F(z)}{z-z_{0}}dz \\ &=& \frac{1}{2 \pi i}\oint_{\gamma}f(z) dz. \end{array}$

As ${F(z_{0}) = 0}$ , it follows that statement ${2}$ implies ${1}$ .

Next, suppose ${z_{0} \not\in \Omega}$ . Then the function ${f(z)=\frac{1}{z-z_{0}}}$ is holomorphic on the set ${\Omega}$ . By statement ${1}$ , we find that