Let $f$ be meromorphic in $\{z\in\mathbb{C}:|z|\le2\}$ with a simple pole at $z=1$ and no other poles. Let $f(z)=\sum_{n=0}^{\infty}a_nz^n$ be the expansion of $f$ valid for $\{z\in\mathbb{C}:|z|<1\}$. Then, can we can we conclude that $\lim_\limits{n\to\infty}a_n=-\text{Res}_{z=1}f(z)$?

I am blown by this problem. I can only think of the definition of residue as $\text{Res}_{z=a}=R$, where $f(z)-\frac{R}{z-a}$ has a derivative. I also think the Laurent series e xpansion has some role to play. Any hints. Thanks beforehand.

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    $\begingroup$ If $f$ has no other poles in $\{ z : \lvert z\rvert < 2\}$, let $c$ be the residue of $f$ at $1$ and consider the function $$g \colon z \mapsto f(z) - \frac{c}{z-1}.$$ What do you know about the function $g$ and the coefficients in its power series expansion about $0$? $\endgroup$ – Daniel Fischer May 4 '17 at 14:32
  • $\begingroup$ @DanielFischer the coefficients of power series expansions of $g$ about $0$ are $(a_n+c)$, i suppose? $\endgroup$ – vidyarthi May 4 '17 at 14:38
  • $\begingroup$ Well, $a_n + c$, since $-\frac{c}{z-1} = \frac{c}{1-z}$. Okay, that's one useful part. What else do you know about the coefficients? (What properties has $g$?) $\endgroup$ – Daniel Fischer May 4 '17 at 14:40
  • $\begingroup$ @DanielFischer ok, so then, just multiply both sides by $(z-1)$ and find the limit at $1$, am i done? $\endgroup$ – vidyarthi May 4 '17 at 14:41
  • $\begingroup$ That gives you the residue, but you want to determine the behaviour of $a_n$ for $n\to \infty$. Can you link those? $\endgroup$ – Daniel Fischer May 4 '17 at 14:44

We need the further assumption that $f$ has no other pole in the closed unit disk for the conclusion. If $f$ has another pole there, we usually don't have $\lim\limits_{n\to\infty} a_n = -\operatorname{Res}(f;1)$.

If that is the case, then $g \colon z \mapsto f(z) - \frac{R}{z-1}$ is holomorphic on a disk with radius $\rho > 1$. Then its MacLaurin series

$$\sum_{n = 0}^{\infty} b_n z^n$$

converges at $z = 1$, whence $\lim\limits_{n\to\infty} b_n = 0$. But

$$g(z) = f(z) + \frac{R}{1-z} = \sum_{n = 0}^{\infty} a_n z^n + R\sum_{n = 0}^{\infty} z^n = \sum_{n = 0}^{\infty} (a_n + R)z^n$$

for $\lvert z\rvert < 1$, and so $b_n = a_n + R$ for all $n$, whence

$$\lim_{n\to \infty} a_n = \lim_{n\to \infty} (b_n - R) = \Bigl(\lim_{n\to\infty} b_n\Bigr) - R = -R.$$

  • $\begingroup$ ok, so the crux of the discussion was convergence of the laurent series in the annulus. Thanks for the discussion. by the way, am I correct in saying that the maclaurin's series for $g$ is the laurent series of $f$? $\endgroup$ – vidyarthi May 4 '17 at 14:54
  • $\begingroup$ We're only looking at power series here (even more restricted, power series about $0$). Of course every power series is also a Laurent series, but it's actually important that we look at power series. Since by assumption $R \neq 0$, the two functions $f$ and $g$ don't coincide on any nonempty open set, so the MacLaurin series of $g$ doesn't represent $f$ anywhere, and hence is not a Laurent series of $f$. The Laurent series of $f$ about $z_0$ would have the form $$\sum_{n = m}^{\infty} c_n(z - z_0)^n,$$ where $m \geqslant 0$ if $f$ is holomorphic at $z_0$ and $m < 0$ if $z_0$ is a pole of $f$. $\endgroup$ – Daniel Fischer May 4 '17 at 15:04

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