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Math 382-0: Kursobjekt

Aaron Greicius

Section 1.22 Poles and residue computation

Definition 1.22.1. Order of isolated singularity.

Let \(z_0\) be an isolated singularity of \(f\) and let \(f(z)=\sum_{n=-\infty}^\infty c_n(z-z_0)^n\) be the \(z_0\)-centered Laurent series expansion in a punctured open ball around \(z_0\text{.}\) We define the order of \(f\) at \(z_0\) (or the order of \(z_0\) with respect to \(f\)), denoted \(\ord_f z_0\) as follows:
  • if there is an integer \(m\) such that \(c_m\ne 0\) and \(c_k=0\) for all integers \(k< m\text{,}\) then \(\ord_f z_0=m\text{;}\)
  • if for all integers \(m\) there exists \(k< m\) such that \(c_m\ne 0\text{,}\) then \(\ord_f z_0=-\infty\text{;}\)
  • if \(c_k=0\) for all integers \(k\text{,}\) then \(\ord_f z_0=\infty\text{.}\)

Definition 1.22.2. Singularity classification.

Let \(z_0\) be an isolated singularity of \(f\text{.}\)
  • If \(\ord_f z_0\geq 0\text{,}\) then \(z_0\) is called a removable singularity of \(f\text{.}\)
  • If \(\ord_f z_0=m\text{,}\) where \(m> 0\) is a positive integer, then \(z_0\) is called a zero of \(f\) of order \(m\).
  • If \(\ord_f z_0=-m\text{,}\) where \(m> 0\) is a positive integer, then \(z_0\) is called a pole of \(f\) of order \(m\).
  • If \(\ord_f z_0=-\infty\text{,}\) then \(z_0\) is called an infinite singularity of \(f\text{.}\)

Example 1.22.3. Singularity classification.

Compute \(\ord_f z_0\) for the given \(f\) and isolated singularity\(z_0\) and classify the singularity \(z_0\text{.}\)
  1. \(f(z)=\frac{\sin z}{z}\text{,}\) \(z_0=0\)
  2. \(f(z)=z^3e^{-1/z}\text{,}\) \(z_0=0\)
  3. \(f(z)=\frac{1}{z^5+z^3}\text{,}\) \(z_0=0\)
Solution.
  1. For \(z\ne 0\text{,}\) we have
    \begin{align*} \frac{\sin z}{z} \amp =\frac{1}{z}(z-z^3/3!+z^5/5!-\cdots)\\ \amp = 1-\frac{1}{6}z^2+\frac{1}{120}z^4-\cdots \text{.} \end{align*}
    We conclude that \(\ord_f 0=0\text{,}\) and hence that \(0\) is a removable singularity.
    The function \(g(z)=1-\frac{1}{6}z^2+\frac{1}{120}-\cdots\) is analytic on \(\C\) and satisfies
    \begin{equation*} g(z)=\begin{cases} \frac{\sin z}{z} \amp \text{if } z\ne 0\\ 1 \amp \text{otherwise} \end{cases}\text{.} \end{equation*}
    In other words, \(g\) is an analytic continuation of \(f\) (analytic on \(\C-\{0\}\)) to \(\C\text{.}\) This illustrates why we call these type of singularities removable: we can extend \(f\) to a function that is also analytic at the singularity in question, thereby “removing” it.
  2. As we have computed previously,
    \begin{align*} z^3e^{1/z}\amp =\sum_{n=0}^\infty \frac{1}{n!} z^{-n+3} \\ \amp = \sum_{n=-1}^{-\infty}\frac{1}{(n+3)! z^{n}}+\frac{1}{6}+\frac{1}{2}z+z^3 \\ \amp = \sum_{n=-\infty}^3 c_nz^n\text{,} \end{align*}
    where \(c_n=\frac{1}{-n+3}!\text{.}\) Since \(c_n\ne 0\) for all \(n\leq 3\text{,}\) we see that \(\ord_f 0=-\infty\text{,}\) and hence that \(0\) is an essential singularity of \(f\text{.}\)
  3. We compute
    \begin{align*} \frac{1}{z^5+z^3} \amp = \frac{1}{z^3}\cdot \frac{1}{z^2+1}\\ \amp = \frac{1}{z^3}(1-z^2+z^4-z^6+\cdots)\\ \amp =\frac{1}{z^3}-\frac{1}{z}+z-z^3+\cdots\text{.} \end{align*}
    Thus \(\ord_f 0=-3\) showing that \(0\) is a pole of order \(3\text{.}\)

Proof.

  1. If \(z_0\) is a removable singularity, then we have
    \begin{equation*} f(z)=a_0+a_1z+\cdots \end{equation*}
    for all \(z\in B_R(z_0)-\{z_0\}\) for some \(R> 0\text{.}\) This power series defines a function \(g\) that is analytic on the entire ball \(B_R(z_0)\text{;}\) since \(g\) is continuous, it is bounded on \(\overline{B}_r(z_0)\) for any \(r< R\text{,}\) and hence also on \(B_r(z_0)\text{.}\) Since \(f(z)=g(z)\) for all \(z\ne z_0\text{,}\) we see that \(f\) is bounded on \(B_r(z_0)-\{z_0\}\text{.}\)
    If \(f\) is bounded on \(B_r(z_0)-\{z_0\}\text{,}\) then there is an \(M\geq 0\) such that \(\abs{f(z)}\leq M\) for all \(z\in B_r(z_0)-\{z_0\}\text{.}\) Letting \(f(z)=\sum_{n=-\infty}^\infty c_n(z-z_0)^n\) be the Laurent series expansion of \(f\) at \(z_0\) for this punctured ball, we have
    \begin{equation*} c_n=\frac{1}{2\pi i}\int_{\gamma_s} \frac{f(z)}{(z-z_0)^{n+1}}\, dz \end{equation*}
    for any \(\gamma(t)=se^{it}\text{,}\) \(t\in [0,2\pi]\) for any \(0< s< r\text{.}\) But then for all such \(s\) and all \(n=-m\) with \(m> 0\text{,}\) we have
    \begin{align*} \abs{c_{n}} \amp = \abs{\frac{1}{2\pi i}\int_{\gamma_s}f(z)(z-z_0)^{m-1}\, dz} \\ \amp \leq \frac{1}{2\pi}M s^{m-1}\cdot 2\pi s \amp (ML-\text{ineq.})\\ \amp = Ms^m \text{.} \end{align*}
    Since \(m\geq 1\text{,}\) we see that \(Ms^m\to 0\) as \(s\to 0\text{.}\) It follows that \(c_{-m}=0\) for all \(m\geq 1\text{.}\) Thus \(z_0\) is a removable singularity.
  2. The proof of this fact is very similar to that of Corollary 1.19.4. As such we will omit some details.
    Suppose \(f(z)=(z-z_0)^mg(z)\) where \(g\) is analytic on \(B_R(z_0)\) and \(g(z_0)\ne 0\text{.}\) Let \(f(z)=\sum_{n=-\infty}^\infty c_n(z-z_0)^n\) be the Laurent series expansion of \(f\) in \(B_R(z_0)-\{z_0\}\text{.}\) Let \(\gamma(t)=z_0+re^{it}\text{,}\) \(t\in [0,2\pi]\text{,}\) for some \(0< r< R\) For all \(k< m\text{,}\) we have
    \begin{align*} c_k \amp = \frac{1}{2\pi i}\int_\gamma\frac{f(z)}{(z-z_0)^{k+1}}\, dz\\ \amp = \frac{1}{2\pi i}\int_\gamma\frac{(z-z_0)^mg(z)}{(z-z_0)^{k+1}}\, dz\\ \amp = \frac{1}{2\pi i}\int_\gamma(z-z_0)^{m-k-1}g(z)\, dz\\ \amp = 0\text{,} \end{align*}
    since \(m-k -1 \geq 0\text{,}\) \(g\) is analytic, and thus \((z-z_0)^{m-k-1}g(z)\) is analytic. Similarly, we have
    \begin{equation*} c_m=\frac{1}{2\pi i}\int_\gamma\frac{g(z)}{z-z_0}\, dz=g(z_0)\ne 0\text{,} \end{equation*}
    using the Cauchy integral formula on the analytic function \(g\text{.}\) This proves that \(\ord_f z_0=m\text{.}\)
    Now assume \(\ord_f z_0=m\text{,}\) so that we have
    \begin{align*} f(z) \amp =a_m(z-z_0)^m+a_{m+1}(z-z_0)^{m+1}+\cdots\\ \amp =(z-z_0)^m(a_m+a_{m+1}(z-z_0)+\cdots )\text{,} \end{align*}
    where we can show that the series \((a_m+a_{m+1}(z-z_0)+\cdots )\) converges on \(B_R(z_0)\) just as in the proof of Corollary 1.19.4, and thus that the function
    \begin{equation*} g(z)=(a_m+a_{m+1}(z-z_0)+\cdots ) \end{equation*}
    is analytic on \(B_R(z_0)\) and satisfies \(g(0)=a_m\ne 0\text{.}\)
  3. We prove the contrapositive. If \(f(B_r(z_0)-\{z_0\})\) is not dense for some \(0< r\lt R\text{,}\) then we can find a \(w\in \C\) and \(\epsilon > 0\) such that \(\abs{f(z)-w)}> \epsilon\) for all \(z\in B_r(z_0)-\{z_0\}\text{.}\) It follows that the function \(g(z)=1/(f(z)-w)\) is analytic on \(U=B_r(z_0)-\{z_0\}\) and satisfies \(\abs{g(z)}< \frac{1}{\epsilon}\) for all \(z\) in this region. It follows from the Riemann removable singularity theorem (part (a)) that \(z_0\) is a removable singularity of \(g(z)\text{,}\) and hence that we may write \(g(z)=a_0+a_1(z-z_0)+\cdots\) for all \(z\in B_r(z_0)-\{z_0\}\text{.}\) Furthermore, since \(a_0=g(z_0)\ne 0\text{,}\) we have
    \begin{equation*} f(z)-w=1/g(z)=b_0+b_1(z-z_0)+b_2(z-z_0)^2+\cdots\text{,} \end{equation*}
    and thus
    \begin{equation*} f(z)=(w+b_0)+b_1(z-z_0)+b_2(z-z_0)^2+\cdots \end{equation*}
    for all \(z\in B_r(z_0)-\{z_0\}\text{.}\) But then \(z_0\) is clearly not an essential singularity of \(f\text{.}\)

Remark 1.22.5. Picard’s theorem.

Remarkably, the Casorati-Weierstrass theorem can be strengthened as follows: if \(z_0\) is an essential singularity of \(f\text{,}\) then for any open punctured ball \(B_r(z_0)-\{z_0\}\) on which \(f\) is analytic, \(f\) attains all values of \(\C\) with one possible exception, infinitely often. This is the content of Picard’s (big) theorem. The necessity of including one possible exception is illustrated by the function \(f(z)=e^{1/z}\text{,}\) which attains all values of \(\C\) except \(0\text{.}\)

Proof.

The proof is left as an exercise.

Proof.

Given the factorization \(f(z)=\frac{g(z)}{(z-z_0)^m}\) as given above, let \(g(z)=\sum_{n=0}^\infty a_n(z-z_0)^n\) be a power series representation of \(g\) at \(z_0\) that converges for all \(z\in B_R(z_0)\) for some \(R\text{.}\) It follows that \(f(z)=g(z)/(z-z_0)^n\) has a Laurent series representation
\begin{equation*} f(z)=\sum_{n=-m}^\infty a_{m+n}(z-z_0)^{n}\text{,} \end{equation*}
and thus that
\begin{equation*} \res_f z_0=a_{m-1}=\frac{g^{(n-1)}(z_0)}{(n-1)!}\text{.} \end{equation*}
The alternative description of \(\res_f z_0\) is obtained by observing that
\begin{equation*} f(z)(z-z_0)^m=g(z)\text{,} \end{equation*}
and thus that
\begin{equation*} g^(n-1)(z)=\frac{d^{n-1}}{dz^{n-1}}\left(f(z)(z-z_0)^m\right) \end{equation*}
for all \(z\in B_R(z_0)-\{z_0\}\text{.}\) Since \(g^{(n-1)}\) is continuous (in fact, infinitely differentiable), we conclude that
\begin{equation*} g^{(n-1)}(z_0)=\lim\limits_{z\to z_0}\frac{d^{n-1}}{dz^{n-1}}\left(f(z)(z-z_0)^m\right)\text{,} \end{equation*}
whence the second equality in (1.82) follows.

Example 1.22.8. Residue computation.

Let \(f(z)=\frac{\cot z}{z}\text{.}\) Find all isolated singularities of \(f\) and compute their residues.
Solution.
Writing \(f(z)=\cos z/(z\sin z)\text{,}\) we see that \(f\) is analytic everywhere except at the points \(z=\pi n\text{,}\) \(n\in \Z\text{.}\) It is easy to see that \(\sin z\) has a simple pole at each \(z_0=n\pi\) for all \(n\text{,}\) and that \(1/z\) has a simple pole at \(z_0=0\text{.}\) Since furthermore \(\cos z\) is nonzero at these points, it follows from our order arithmetic corollary that
\begin{equation*} \ord_f (n\pi)=\begin{cases} -1 \amp \text{if } n\ne 0\\ -2 \amp \text{if } n=0 \end{cases}\text{.} \end{equation*}
For \(z_0=n\pi\) with \(n\ne 0\text{,}\) we conclude, using Theorem 1.21.7 that
\begin{align*} \res_f z_0 \amp = \frac{\cos z_0}{\frac{d}{dz}(z\sin z)\vert_{z=z_0}}\\ \amp = \frac{\cos(n\pi)}{\sin(n\pi)+(n\pi)\cos(n\pi)}\\ \amp = \frac{1}{n\pi}\text{.} \end{align*}
For \(z_0=0\text{,}\) we have
\begin{align*} \frac{\cos z}{z\sin z} \amp =\frac{\cos z}{z^2(1-z^2/6+z^4/120-\cdots}\\ \amp = \frac{\cos z}{z^2g(z)} \end{align*}
where \(g(z)=1-z^2/6+z^4/120-\cdots\text{.}\) Since \(h(z)=\cos z/g(z)\) is analytic about \(0\) we have
\begin{equation*} h(z)=\cos z/g(z)=a_0+a_1z+\cdots \text{,} \end{equation*}
and thus
\begin{equation*} f(z)=\frac{1}{z^2}h(z)=\frac{a_0}{z^2}+\frac{a_1}{z}+\cdots\text{.} \end{equation*}
Thus \(\res_f 0=a_1\text{.}\) Lastly, since \(a_n=\frac{h^{(n)}(0)}{n!}\text{,}\) we see that
\begin{equation*} a_1=h'(0)=\frac{\sin(0)g(0)-\cos(0)g'(0)}{g(0)^2}=\frac{0-0}{1}=0\text{.} \end{equation*}
We conclude that
\begin{equation*} \res_f 0=a_1=0\text{.} \end{equation*}
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