Question 7.33: Prove Mittag –Leffler’s expansion theorem (see page 209).
Prove Mittag –Leffler’s expansion theorem (see page 209).
Let f(z) have poles at z=a_{n}, n=1,2, \ldots, and suppose that z=\zeta is not a pole of f(z). Then, the function f(z) / z-\zeta has poles at z=a_{n}, n=1,2,3, \ldots and \zeta.
Residue of f(z) / z-\zeta at z=a_{n}, n=1,2,3, \ldots, is
\underset{z \rightarrow a_{n}}{\lim}\left(z-a_{n}\right) \frac{f(z)}{z-\zeta}=\frac{b_{n}}{a_{n}-\zeta}
Residue of f(z) / z-\zeta at z=\zeta is
\underset{z \rightarrow \zeta}{\lim}(z-\zeta) \frac{f(z)}{z-\zeta}=f(\zeta)
Then, by the residue theorem,
\frac{1}{2 \pi i} \oint\limits_{C_{N}} \frac{f(z)}{z-\zeta} d z=f(\zeta)+\sum\limits_{n} \frac{b_{n}}{a_{n}-\zeta} (1)
where the last summation is taken over all poles inside circle C_{N} of radius R_{N} (Fig. 7-14). Suppose that f(z) is analytic at z=0. Then, putting \zeta=0 in (1), we have
\frac{1}{2 \pi i} \oint\limits_{C_{N}} \frac{f(z)}{z} d z=f(0)+\sum\limits_{n} \frac{b_{n}}{a_{n}} (2)
Subtraction of (2) from (1) yields
\begin{aligned} f(\zeta)-f(0)+\sum\limits_{n} b_{n}\left(\frac{1}{a_{n}-\zeta}-\frac{1}{a_{n}}\right) & =\frac{1}{2 \pi i} \oint\limits_{C_{N}} f(z)\left\{\frac{1}{z-\zeta}-\frac{1}{z}\right\} d z \\ & =\frac{\zeta}{2 \pi i} \oint\limits_{C_{N}} \frac{f(z)}{z(z-\zeta)} d z \qquad (3) \end{aligned}
Now since |z-\zeta| \geq|z|-|\zeta|=R_{N}-|\zeta| for z on C_{N}, we have, if |f(z)| \leq M,
\left|\oint\limits_{C_{N}} \frac{f(z)}{z(z-\zeta)} d z\right| \leq \frac{M \cdot 2 \pi R_{N}}{R_{N}\left(R_{N}-|\zeta|\right)}
As N \rightarrow \infty and therefore R_{N} \rightarrow \infty, it follows that the integral on the left approaches zero, i.e.,
\underset{N \rightarrow \infty}{\lim} \oint\limits_{C_{N}} \frac{f(z)}{z(z-\zeta)} d z=0
Hence from (3), letting N \rightarrow \infty, we have as required
f(\zeta)=f(0)+\sum\limits_{n} b_{n}\left(\frac{1}{\zeta-a_{n}}+\frac{1}{a_{n}}\right)
the result on page 209 being obtained on replacing \zeta by z.
