Question 6.25: Prove Laurent’s theorem: Suppose f(z) is analytic inside and...
Prove Laurent’s theorem: Suppose f(z) is analytic inside and on the boundary of the ring-shaped region \mathcal{R} bounded by two concentric circles C_{1} and C_{2} with center at a and respective radii r_{1} and r_{2}\left(r_{1}>r_{2}\right) (see Fig. 6-5). Then for all z in \mathcal{R},
f(z)=\sum\limits_{n=0}^{\infty} a_{n}(z-a)^{n}+\sum\limits_{n=1}^{\infty} \frac{a-n}{(z-a)^{n}}
where
\begin{aligned} & a_{n}=\frac{1}{2 \pi i} \oint\limits_{C_{1}} \frac{f(w)}{(w-a)^{n+1}} d w \quad n=0,1,2, \ldots \\ & a_{-n}=\frac{1}{2 \pi i} \oint\limits_{C_{2}} \frac{f(w)}{(w-a)^{-n+1}} d w \quad n=1,2,3, \ldots \end{aligned}
By Cauchy’s integral formula [see Problem 5.23, page 159], we have
f(z)=\frac{1}{2 \pi i} \oint\limits_{C_{1}} \frac{f(w)}{w-z} d w-\frac{1}{2 \pi i} \oint\limits_{C_{2}} \frac{f(w)}{w-z} d w (1)
Consider the first integral in (1). As in Problem 6.22, equation (2), we have
\begin{aligned} \frac{1}{w-z} & =\frac{1}{(w-a)\{1-(z-a) /(w-a)\}} \\ & =\frac{1}{w-a}+\frac{z-a}{(w-a)^{2}}+\cdots+\frac{(z-a)^{n-1}}{(w-a)^{n}}+\left(\frac{z-a}{w-a}\right)^{n} \frac{1}{w-z} \quad (2) \end{aligned}
so that
\begin{aligned} \frac{1}{2 \pi i} \oint\limits_{C_{1}} \frac{f(w)}{w-z} d w= & \frac{1}{2 \pi i} \oint\limits_{C_{1}} \frac{f(w)}{w-a} d w+\frac{z-a}{2 \pi i} \oint\limits_{C_{1}} \frac{f(w)}{(w-a)^{2}} d w \\ & +\cdots+\frac{(z-a)^{n-1}}{2 \pi i} \oint\limits_{C_{1}} \frac{f(w)}{(w-a)^{n}} d w+U_{n} \\ = & a_{0}+a_{1}(z-a)+\cdots+a_{n-1}(z-a)^{n-1}+U_{n} \quad (3) \end{aligned}
where
a_{0}=\frac{1}{2 \pi i} \oint\limits_{C_{1}} \frac{f(w)}{w-a} d w, \quad a_{1}=\frac{1}{2 \pi i} \oint\limits_{C_{1}} \frac{f(w)}{(w-a)^{2}} d w, \quad \ldots, \quad a_{n-1}=\frac{1}{2 \pi i} \oint\limits_{C_{1}} \frac{f(w)}{(w-a)^{n}} d w
and
U_{n}=\frac{1}{2 \pi i} \oint\limits_{C_{1}}\left(\frac{z-a}{w-a}\right)^{n} \frac{f(w)}{w-z} d w
Let us now consider the second integral in (1). We have on interchanging w and z in (2),
\begin{aligned} -\frac{1}{w-z} & =\frac{1}{(z-a)\{1-(w-a) /(z-a)\}} \\ & =\frac{1}{z-a}+\frac{w-a}{(z-a)^{2}}+\cdots+\frac{(w-a)^{n-1}}{(z-a)^{n}}+\left(\frac{w-a}{z-a}\right)^{n} \frac{1}{z-w} \end{aligned}
so that
\begin{aligned} -\frac{1}{2 \pi i} \oint\limits_{C_{2}} \frac{f(w)}{w-z} d w= & \frac{1}{2 \pi i} \oint\limits_{C_{2}} \frac{f(w)}{z-a} d w+\frac{1}{2 \pi i} \oint\limits_{C_{2}} \frac{w-a}{(z-a)^{2}} f(w) d w \\ & +\cdots+\frac{1}{2 \pi i} \oint\limits_{C_{3}} \frac{(w-a)^{n-1}}{(z-a)^{n}} f(w) d w+V_{n} \\ = & \frac{a_{-1}}{z-a}+\frac{a_{-2}}{(z-a)^{2}}+\cdots+\frac{a_{-n}}{(z-a)^{n}}+V_{n} \quad (4) \end{aligned}
where
a_{-1}=\frac{1}{2 \pi i} \oint\limits_{C_{2}} f(w) d w, \quad a_{-2}=\frac{1}{2 \pi i} \oint\limits_{C_{2}}(w-a) f(w) d w, \ldots, a_{-n}=\frac{1}{2 \pi i} \oint\limits_{C_{2}}(w-a)^{n-1} f(w) d w
and
V_{n}=\frac{1}{2 \pi i} \oint\limits_{C_{2}}\left(\frac{w-a}{z-a}\right)^{n} \frac{f(w)}{z-w} d w
From (1), (3), and (4), we have
\begin{aligned} f(z)= & \left\{a_{0}+a_{1}(z-a)+\cdots+a_{n-1}(z-a)^{n-1}\right\} \\ & +\left\{\frac{a_{-1}}{z-a}+\frac{a_{-2}}{(z-a)^{2}}+\cdots+\frac{a_{-n}}{(z-a)^{n}}\right\}+U_{n}+V_{n} \end{aligned} (5)
The required result follows if we can show that (a) \lim _{n \rightarrow \infty} U_{n}=0 and (b) \lim _{n \rightarrow \infty} V_{n}=0. The proof of (a) follows from Problem 6.22. To prove (b), we first note that since w is on C_{2},
\left|\frac{w-a}{z-a}\right|=\kappa<1
where \kappa is a constant. Also, we have |f(w)|<M where M is a constant and
|z-w|=|(z-a)-(w-a)| \geq|z-a|-r_{2}
Hence, from Property (e), page 112, we have
\begin{aligned} \left|V_{n}\right| & =\frac{1}{2 \pi}\left|\oint\limits_{C_{2}}\left(\frac{w-a}{z-a}\right)^{n} \frac{f(w)}{z-w} d w\right| \\ & \leq \frac{1}{2 \pi} \frac{\kappa^{n} M}{|z-a|-r_{2}} 2 \pi r_{2}=\frac{\kappa^{n} M r_{2}}{|z-a|-r_{2}} \end{aligned}
Then, \lim _{n \rightarrow \infty} V_{n}=0 and the proof is complete.