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Math 382-0: Kursobjekt

Aaron Greicius

Section 1.20 Laurent series

We have begun to see the utility of power series representations for functions \(f\) that are holomorphic on an open ball \(B_R(z_0)\text{.}\) It turns out that much of this theory (and its attendant benefits) can be extended to more general functions by relaxing somewhat the differentiability condition. Namely, we will develop a series representation for functions \(f\) which are holomorphic not on a full open ball centered at \(z_0\text{,}\) but only on an annulus of the form \(\{z\in \C\colon R_1< \abs{z-z_0}< R_2\}\text{.}\) We proceed directly to Theorem 1.20.1, which will serve to motivate our study of Laurent series.

Proof.

The proof is very simple to the proof of Theorem 1.18.1: the main fact we use is the Cauchy integral formula applied to elements within \(A\text{.}\) In more detail, given \(w\in A\text{,}\) pick a closed ball \(\overline{B}_s(z_0)\subseteq A\text{,}\) and let \(\phi=w+se^{it}\text{,}\) \(t\in [0,2\pi]\text{.}\) Next pick \(R\) and \(S\) satisfying \(R_1< R < S < R_2\) such that \(\overline{B}_s(z_0)\) lies in the annular region \(R< \abs{z-z_0}< S\text{,}\) and let \(\gamma_R(t)=z_0+Re^{it}\) and \(\gamma_S(t)=z_0+Se^{it}\text{,}\) \(t\in [0,2\pi]\text{.}\) (See Figure 1.20.2.) By the Cauchy integral formula, we have
\begin{equation*} f(w)=\frac{1}{2\pi i}\int_\phi \frac{f(z)}{z-w}\, dz\text{.} \end{equation*}
By the principle of path deformation we have
\begin{equation*} f(w)=\frac{1}{2\pi i}\int_\phi \frac{f(z)}{z-w}\, dz=\frac{1}{2\pi i}\int_{\gamma_S}\frac{f(z)}{z-w}\, dz-\frac{1}{2\pi i}\int_{\gamma_R}\frac{f(z)}{z-w}\, dz \text{.} \end{equation*}
Since \(\abs{(w-z_0)/(z-w_0)}< 1\) for all \(z\in \gamma_S\text{,}\) we have
\begin{align*} \frac{1}{2\pi i}\int_{\gamma_S}\frac{f(z)}{z-w}\, dw \amp = \frac{1}{2\pi i}\int_{\gamma_S}\frac{f(z)}{z-z_0}\cdot \frac{1}{1-(w-z_0)/(z-z_0)}\, dw \\ \amp = \frac{1}{2\pi i}\int_{\gamma_S}(\frac{f(z)}{z-z_0}\sum_{n=0}^\infty((w-z_0)/(z-z_0))^n)\, dz\\ \amp =\frac{1}{2\pi i} \sum_{n=0}^\infty\int_{\gamma_S}(\frac{f(z)}{(z-z_0)^{n+1}}\, dz)(w-z_0)^n\\ \amp = \sum_{n=0}^\infty a_n(w-z_0)^n , \ a_n=\frac{1}{2\pi i}\int_{\gamma_S}\frac{f(z)}{(z-z_0)^n}\, dz\text{.} \end{align*}
Similarly, since \(\abs{(z-z_0)/(w-z_0)}< 1\) for all \(z\in \gamma_R\text{,}\) we have
\begin{align*} \frac{1}{2\pi i}\int_{\gamma_R}\frac{f(z)}{z-w}\, dw \amp = \frac{1}{2\pi i}\int_{\gamma_R}\frac{f(z)}{w-z_0}\cdot \frac{-1}{1-(z-z_0)/(w-z_0)}\, dw \\ \amp = -\frac{1}{2\pi i}\int_{\gamma_R}(\sum_{n=0}^\infty\frac{f(z)(z-z_0)^{n}}{(w-z_0)^{n+1}})\, dz\\ \amp = -\frac{1}{2\pi i}\sum_{n=0}^\infty\int_{\gamma_R}(\frac{f(z)}{(z-z_0)^{n}}\, dz)(w-z_0)^{n+1}\\ \amp = -\sum_{n=1}^\infty b_n(w-z_0)^{-n}, \ b_n=\frac{1}{2\pi i}\int_{\gamma_R}f(z)(z-z_0)^{n-1}\, dz\text{.} \end{align*}
(Note that in both these computations we are justified in bringing the integral into the infinite sum thanks to the uniform convergence of this sum to the given function.)
Putting it all together, we have
\begin{align*} f(w)\amp =\frac{1}{2\pi i}\int_{\gamma_S}\frac{f(z)}{z-w}\, dz-\frac{1}{2\pi i}\int_{\gamma_R}\frac{f(z)}{z-w}\, dz\\ \amp = \sum_{n=0}^\infty a_n(w-z_0)^n+\sum_{n=1}^\infty \frac{b_n}{(w-z_0)^n} \end{align*}
for all \(w\in A\text{,}\) where (using the principle of deformation once again)
\begin{align*} a_n \amp=\frac{1}{2\pi i}\int_{\gamma_S}\frac{f(z)}{(z-z_0){n+1}}\, dz=\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{(z-z_0){n+1}}\, dz \\ b_n \amp=\frac{1}{2\pi i}\int_{\gamma_R}f(z)(z-z_0)^{n-1}\, dz=\frac{1}{2\pi i}\int_{\gamma}f(z)(z-z_0)^{n-1}\, dz \text{.} \end{align*}
Recall that \(\gamma(t)=z_0+re^{it}\text{,}\) \(t\in [0,2\pi]\) for some \(r\in (R_1,R_2)\text{.}\)
Visual idea behind proof of theorem
Figure 1.20.2. Visual idea behind proof of Theorem 1.20.1

Definition 1.20.3. Laurent series.

Given \(z_0\in \C\text{,}\) a (complex) laurent series centered at \(z_0\) in the variable \(z\) is an expression of the form
\begin{equation} f(z)=\sum_{n=-\infty}^\infty c_n(z-z_0)^n= \cdots \frac{c_{-2}}{(z-z_0)^2}+\frac{c_{-1}}{z-z_0}+c_0+c_1(z-z_0)+c_2(z-z_0)^2+\cdots\text{,}\tag{1.77} \end{equation}
where \(c_n\in \C\text{.}\)
Let
\begin{align} g(u) \amp =\sum_{n=0}^\infty a_n u^n, a_n=c_n\tag{1.78}\\ h(u) \amp =\sum_{n=1}^\infty b_n u^n, b_n=c_{-n}\tag{1.79} \end{align}
so that
\begin{equation*} f(z)=g(z-z_0)+h(1/(z-z_0))\text{.} \end{equation*}
We say the Laurent series \(f\) converges at a complex number \(z\) if the series \(g(z-z_0)\) and \(h(1/(z-z_0))\) both converge, and diverges otherwise. In the case of convergence, we call \(g(z-z_0)+h(1/(z-z_0))\) the value of the Laurent series.

Example 1.20.4. Laurent series expansion.

Let \(f(z)=z^3e^{1/z}\text{.}\) Find a Laurent series expansion of \(f\) centered at \(z_0=0\) in the annulus
\begin{equation*} A=\{z\in \C\colon 0< \abs{z} < \infty\}=\C-\{0\}\text{.} \end{equation*}
Solution.
Since the power series representation
\begin{equation*} e^{z}=\sum_{n=0}^\infty \frac{z^n}{n!} \end{equation*}
is valid for all \(z\text{,}\) we have
\begin{equation*} e^{1/z}=\sum_{n=0}^\infty \frac{1}{n! z^n} \end{equation*}
and
\begin{align*} z^3e^{1/z} \amp =\sum_{n=0}^\infty \frac{1}{n!z^{n+3}}\\ \amp =z^3+z^2+\frac{1}{2}z+\frac{1}{6}+\frac{1}{24z}+\cdots \\ \amp = \sum_{n=-1}^\infty \frac{1}{(-n+3)!}z^{n}+ \frac{1}{2}+z^2+z^3 \end{align*}
for all \(z\in \C-\{0\}\text{.}\)

Remark 1.20.5. Laurent series.

Since the notion of convergence of a Laurent series boils down to the convergence of the two power series (1.78)–(1.79), Laurent series inherit many of the useful properties of power series. For example, from the uniform convergence properties of \(g\) and \(h\text{,}\) it follows that the Laurent series \(f\) can be differentiated and integrated term-wise, just as with power series.
As we saw in Theorem 1.20.1, however, the natural region of convergence of a Laurent series is an open annulus, as opposed to an open ball. Indeed, if the radii of convergence for \(g\) and \(h\) are \(R\) and \(S\text{,}\) respectively, then \(g\) converges at all \(z\) with \(\abs{z-z_0}< R\) and \(h\) converges for all \(z\) with \(\abs{1/(z-z_0)}< S\text{.}\) Letting \(R'=1/S\text{,}\) it follows that \(f\) converges for all \(z\) satisfying \(R' < \abs{z-z_0}< R\text{,}\) and diverges for all \(z\) satisfying \(\abs{z-z_0}< R'\) or \(\abs{z-z_0}> R\text{.}\) Accordingly, we call \(A=\{z\in \C\colon R' < \abs{z-z_0}< R\}\) the annulus of convergence of \(f\text{.}\)
As a consequence of the discussion above, given a Laurent series, there is a well-defined annulus of convergence associated to it. As the next example illustrates, however, if we begin with a function \(f\) and seek out Laurent series expansions of \(f\text{,}\) these expansions depend on the annulus we choose.

Example 1.20.6. Different Laurent series.

Let \(f(z)=\frac{1}{z^2-4z+3}\text{.}\) Compute a convergent Laurent series for \(f\) centered at \(z_0=-1\) in the given annulus \(A\)
  1. \(\displaystyle A=\{z\in \C\colon 0< \abs{z+1} < 2\}\)
  2. \(\displaystyle A=\{z\in \C\colon 2< \abs{z+1} < 4\}\)
  3. \(\displaystyle A=\{z\in \C\colon 4< \abs{z+1} <\infty\}\)
Solution.
We first use partial fractions to write
\begin{equation*} f(z)=\frac{1}{(z-1)(z-3)}=-\frac{1}{2}\left(\frac{1}{z-1}-\frac{1}{z-3}\right)\text{.} \end{equation*}
Each computation below then boils down to deriving a Laurent series expansion of \(1/(z-1)\) and \(1/(z-3)\) in the given region.
  1. If \(0< \abs{z+1} < \abs{2}\) then \(\abs{(z+1)/2}< 1\) and \(\abs{(z+1)/4}< 1\text{.}\) We maneuver into a place where we can use the geometric series for both fractions:
    \begin{align*} \frac{1}{z-1} \amp = \frac{1}{(z+1)-2}\\ \amp = -\frac{1}{2}\frac{1}{1-(z+1)/2}\\ \amp = -\frac{1}{2}\sum_{n=0}^\infty \frac{1}{2^n}(z+1)^n\\ \frac{1}{z-3} \amp = \frac{1}{(z+1)-4}\\ \amp = -\frac{1}{4}\frac{1}{1-(z+1)/4}\\ \amp = -\frac{1}{4}\sum_{n=0}^\infty \frac{1}{4^n}(z+1)^n \end{align*}
    We conclude that
    \begin{align*} f(z) \amp = -\frac{1}{2}\left(\frac{1}{z-1}-\frac{1}{z-3}\right) \\ \amp = \frac{1}{4}\sum_{n=0}^\infty \frac{1}{2^n}(z+1)^n-\frac{1}{8}\sum_{n=0}\frac{1}{4^n}(z+1)^n\\ \amp = \frac{1}{2}\sum_{n=0}^\infty \frac{2^{n+1}-1}{4^{n+1}}(z+1)^n \text{.} \end{align*}
    It should come as no surprise that our Laurent series expansion in this case is a power series, as \(f\) is in fact analytic on the entire ball \(\abs{z+1}< 2\text{.}\)
  2. For \(z\) satisfying \(2< \abs{z+1} < 4\text{,}\) we may continue to use our power series expansion for the second fraction, but not the first, since now \(\abs{(z+1)/2}> 1\text{.}\) But of course, this is equivalent to \(\abs{2/(z+1)}< 1\text{.}\) So we maneuver accordingly:
    \begin{align*} \frac{1}{z-1} \amp = \frac{1}{(z+1)-2}\\ \amp = \frac{1}{z+1}\frac{1}{1-2/(z+1)}\\ \amp = \sum_{n=0}^\infty \frac{2^n}{(z+1)^{n+1}}\\ \frac{1}{z-3} \amp = -\frac{1}{4}\sum_{n=0}^\infty \frac{1}{4^n}(z+1)^n \end{align*}
    Thus
    \begin{align*} f(z)\amp =-\frac{1}{2}\left(\frac{1}{z-1}-\frac{1}{z-3}\right) \\ \amp = -\frac{1}{2}\sum_{n=0}^\infty \frac{2^n}{(z+1)^{n+1}} -\frac{1}{8}\sum_{n=0}^\infty \frac{1}{4^n}(z+1)^n \\ \amp = -\sum_{n=0}^\infty \frac{2^{n-1}}{(z+1)^{n+1}} -\frac{1}{8}\sum_{n=0}^\infty \frac{1}{4^n}(z+1)^n \\ \amp = -\sum_{n=1}^\infty \frac{2^{n-2}}{(z+1)^{n}} -\frac{1}{8}\sum_{n=0}^\infty \frac{1}{4^n}(z+1)^n \text{.} \end{align*}
  3. Finally, for \(\abs{z+1}> 4\text{,}\) we have \(\abs{2/(z+1)}< 1\) and \(\abs{4/(z+1)}< 1\text{,}\) yielding (similarly as above)
    \begin{align*} \frac{1}{z-1} \amp = \sum_{n=0}^\infty \frac{2^n}{(z+1)^{n+1}}\\ \frac{1}{z-3} \amp = \sum_{n=0}^\infty\frac{4^n}{(z+1)^{n+1}} \end{align*}
    and thus
    \begin{align*} f(z) \amp = \frac{1}{2}\sum_{n=0}^\infty\frac{4^n}{(z+1)^{n+1}} - \frac{1}{2}\sum_{n=0}^\infty \frac{2^n}{(z+1)^{n+1}}\\ \amp = \frac{1}{2}\sum_{n=0}^\infty \frac{4^n-2^n}{(z+1)^{n+1}}\\ \amp = \sum_{n=1}^\infty \frac{2^{n-2}(2^{n-1}-1)}{(z+1)^n}\text{.} \end{align*}

Remark 1.20.7. Multiplying power series.

The next example requires multiplying together two power series. In general, given two power series
\begin{align*} f(z) \amp = \sum_{n=0}^\infty a_n(z-z_0)^n \amp g(z)\amp =\sum_{n=0}^\infty b_n(z-z_0)^n \end{align*}
that converge for all \(z\in B_R(z_0)\text{,}\) their product has power series expansion
\begin{equation*} f(z)g(z)=\sum_{n=0}^\infty c_n(z-z_0)^n\text{,} \end{equation*}
where
\begin{equation*} c_n=a_0b_n+a_1b_{n-1}+\cdots a_nb_0=\sum_{k=0}^na_kb_{n-k}\text{.} \end{equation*}

Example 1.20.8. Truncated Laurent series.

Let \(A\) be the annulus \(0< \abs{z} < 1\text{.}\) Compute the Laurent series \(\sum_{n=-\infty}^\infty c_n z^n\) of \(f(z)=e^z/(z^3+z^2)\) in \(A\) up through \(c_2\text{.}\)
Solution.
We have
\begin{align*} f(z) \amp =\frac{e^z}{z^2(z+1)}\\ \amp = \frac{1}{z^2}\cdot (1+z+\frac{z^2}{2}+\frac{z^3}{6}+\cdots)(1-z+z^2-z^3+\cdots)\\ \amp = \frac{1}{z^2}\sum_{n=0}^\infty c_n z^n, \ c_n=\sum_{k=0}^n(-1)^{n-k}/n!\\ \amp = \frac{1}{z^2}(1+0z+\frac{1}{2}z^2-\frac{1}{3}z^3+\frac{3}{8}z^4+\cdots)\\ \amp = \frac{1}{z^2}+\frac{1}{2}-\frac{1}{3}z+\frac{3}{8}z^2+\cdots \end{align*}

Remark 1.20.9. Expanding a polynomial about a different center.

Since a polynomial \(f(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots +a_1z+a_0\) is entire, we can expand it as a power series about any center \(z_0\text{.}\) We see using formula (1.72) that
\begin{align*} f(z) \amp = \sum_{k=0}^\infty c_k(z-z_0)^k \\ \amp = \sum_{k=0}^\infty \frac{f^{(k)}(z_0)}{k!}(z-z_0)^k\\ \amp = f(z_0)+f'(z_0)(z-z_0)+\frac{f''(z_0)}{2}(z-z_0)^2+\cdots +\frac{f^{(n)}(z_0)}{n!}(z-z_0)^n\text{,} \end{align*}
since \(f^{(k)}\) is the zero function for all \(k\geq n+1\text{.}\)

Example 1.20.10. Laurent series of rational function.

Let \(A\) be the annulus \(0< \abs{z-1} < \infty\text{:}\) i.e., \(A=\C-\{1\}\text{.}\) Compute the Laurent series of \(f(z)=(z^3+2z^2)/(z-1)^3\) on \(A\text{.}\)
Solution.
Let \(g(z)=z^3+2z^2\text{.}\) We first expand \(g\) about \(z_0=1\) as
\begin{align*} g(z)\amp =g(1)+g'(1)(z-1)+\frac{g''(0)}{2!}(z-1)^2+\frac{g'''(0)}{3!}(z-1)^3\\ \amp =3+7(z-1)+5(z-1)^2+(z-1)^3\amp \text{.} \end{align*}
Now compute
\begin{align*} f(z) \amp = \frac{g(z)}{(z-1)^3}=\frac{3}{(z-1)^3}+\frac{7}{(z-1)^2}+\frac{5}{z-1}+1\text{.} \end{align*}
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