0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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: $$ \begin{align} {\rm Res}\left(\frac{\phi}{\psi},a\right)&= \phi'(a)b_0 + \phi(a) b_1 \\ &= \phi'(a)\frac{2}{\psi''(a)} + \phi(a) \left(-\frac{2\psi'''(a)}{3(\psi''(a))^2}\right) \\ &= \frac{2\phi'(a)}{\psi''(a)} - \frac{2\phi(a)\psi'''(a)}{3(\psi''(a))^2} \\ &= \frac{6\phi'(a)\psi''(a)-2\phi(a)\psi'''(a)}{(\psi''(a))^2}.\end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.