Let $\phi$ and $\psi$ be holomorphic functions around $z = a$, where $\phi(a) \neq 0$ and $a$ is a double root of $\psi(z) = 0$. Prove that the residue of $\phi(z)/\psi(z)$ at $z = a$ is: $$\frac{6\phi'(a)\psi''(a)-2\phi(a)\psi'''(a)}{(\psi''(a))^2}.$$

I understand that if $a$ is a pole of order $k$ of a function $f$, then ${\rm Res}(f,a) = g^{(k-1)}(a)/(k-1)!$, where $g(z) = (z-a)^kf(z)$. Meaning, we cover the pole before doing the computation. If $a$ is a double root of $\psi(z)=0$, I understand that $a$ is a double pole of $\psi$, so by the above we should get: $${\rm Res}\left(\frac{\phi}{\psi},a\right) = \frac{\rm d}{{\rm d}z}\Bigg|_{z=a}\left(\frac{(z-a)^2\phi(z)}{\psi(z)}\right).$$Well, crap, this evaluates to zero. What am I missing here?

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    $\begingroup$ Doesn't it evaluate directly to an indeterminate form $0/0$ when $z = a$? I would assume that once you take the derivative, you then have to take the appropriate limit. $\endgroup$ – Michael Seifert Jun 19 '15 at 17:07
  • $\begingroup$ Oh. I think you're right, I messed up. I'll try again here. $\endgroup$ – Ivo Terek Jun 19 '15 at 17:08

Probly what you're doing is just differentiating that fraction by the quotient rule, then setting $z=a$ and noticing the numerator vanishes? Oops, the denominator vanishes too...

It's not quite right to say $g(z)=(z-a)^kf(z)$. At least not in the sense you're doing it. You define $g$ by that formula for $z\ne a$, and then note that $g$ has a removable singularity at $a$.

So what it comes to is instead of saying the residue is the derivative of that fraction at $z=a$, say the residue is the limit as $z\to a$ of the derivative of that fraction. See how that works.

Btw, a sort of different approach that often comes out cleaner: Since $\psi$ has a double root you can write $\psi(z)=(z-a)^2h(z)$, where $h$ is holomorphic. Use that together with the power series for $\phi$...

OK, since I don't see anything else to do right now, let's see if I can take this a step farther. Let's say $\psi=(z-a)^2h(z)$; now $h$ is holomorphic and $h(a)\ne 0$. Now for $z\ne a$ we have $$(z-a)^2f(z)=\phi(z)/h(z).$$But the good news is that that formula remains valid for $z=a$. We get $$g'(a)=(\phi'(a)h(a)-\phi(a)h'(a))/h(a)^2.$$

Figure out what $h(a)$ has to so with $\psi''(a)$; they're almost the same. Also figure out what $h'(a)$ has to so with $\psi''(a)$; here might be a good place for power series: Say $h(z) = c + d(z-a) + ...$. Then $\psi(z)=c(z-a)^2+d(z-a)^3+...$ So $\psi'''(a)$ is something times $d$, which is turn is something times $h'(a)$. Should work out...

  • $\begingroup$ Yes, you're right.. I tried to compute the limit using L'Hospital's rule but that's insane. I'm not sure of how to use the series for $\phi$, though. $\endgroup$ – Ivo Terek Jun 19 '15 at 17:23
  • $\begingroup$ Gawd no don't ever use L'Hospital for anything! I added some detail in an edit just now (turned out a different power series was the one that seemed useful) $\endgroup$ – David C. Ullrich Jun 19 '15 at 18:21

I managed to solve the problem after talking to the TA here, and I'll leave my solution, for completeness.

If $\phi$ and $\psi$ are holomorphic around $a$, in a sufficient small ball around $a$ we have: $$\phi(z) = \sum_{n \geq 0}\frac{\phi^{(n)}(a)}{n!}(z-a)^n, \quad \psi(z) = \sum_{n \geq 0} \frac{\psi^{(n+2)}(a)}{(n+2)!}(z-a)^{n+2},$$because $a$ is a double zero of $\psi$. We can write $\psi(z) = (z-a)^2h(z)$ with $h$ holomorphic around $a$ and $h(a) \neq 0$. This expression gives: $$h(z) = \sum_{n \geq 0} \frac{\psi^{(n+2)}(a)}{(n+2)!}(z-a)^{n}.$$

Since $h(a) \neq 0$ we can write: $$\frac{1}{h(z)} = \sum_{n \geq 0}b_n(z-a)^n,$$for some coefficients $b_n$. I'm writing it like this, so we only compute what we need, later. We have: $$\begin{align} \frac{\phi(z)}{\psi(z)} &= \frac{\phi(z)}{(z-a)^2h(z)} \\ &= \frac{1}{(z-a)^2}\phi(z) \frac{1}{h(z)} \\ &= \frac{1}{(z-a)^2}\left(\sum_{n \geq 0}\frac{\phi^{(n)}(a)}{n!}(z-a)^n\right)\left(\sum_{n \geq 0}b_n(z-a)^n\right) \\ &= \frac{1}{(z-a)^2} \sum_{n \geq 0}\left(\sum_{k=0}^n \frac{\phi^{(n-k)}(a)}{(n-k)!}b_k\right)(z-a)^n \\ &= \sum_{n \geq 0}\left(\sum_{k=0}^n \frac{\phi^{(n-k)}(a)}{(n-k)!}b_k\right)(z-a)^{n-2}\end{align}$$We get the residue we want for $n=1$: $${\rm Res}\left(\frac{\phi}{\psi},a\right) = \sum_{k=0}^1 \frac{\phi^{(1-k)}(a)}{(1-k)!}b_k = \phi'(a)b_0 + \phi(a) b_1.$$So we compute $b_0$ and $b_1$ from the expression for $h$, using the Cauchy product... $$\frac{\psi''(a)}{2}b_0 = 1 \implies b_0 = \frac{2}{\psi''(a)},$$and: $$\frac{\psi''(a)}{2}b_1 + \frac{\psi'''(a)}{3!}\frac{2}{\psi''(a)}=0 \implies b_1 = -\frac{2\psi'''(a)}{3(\psi''(a))^2}.$$Finally: $$ \begin{align} {\rm Res}\left(\frac{\phi}{\psi},a\right)&= \phi'(a)b_0 + \phi(a) b_1 \\ &= \phi'(a)\frac{2}{\psi''(a)} + \phi(a) \left(-\frac{2\psi'''(a)}{3(\psi''(a))^2}\right) \\ &= \frac{2\phi'(a)}{\psi''(a)} - \frac{2\phi(a)\psi'''(a)}{3(\psi''(a))^2} \\ &= \frac{6\phi'(a)\psi''(a)-2\phi(a)\psi'''(a)}{(\psi''(a))^2}.\end{align}$$


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