Proof of the residue formula Res$(f(z);z_0) = -\frac{q''(z_0)}{(q'(z_0))^3}$ for a pole of order two

Question

Suppose $f(z) = \frac{1}{(q(z))^2}$ where $q(z_0) = 0$ and $q'(z_0) \neq 0$

Show that Res$(f(z);z_0) = -\frac{q''(z_0)}{(q'(z_0))^3}$

I been messing around with $f$ for quite a while and honestly I have no idea how to even start.

Any help or insights (or point me in the right direction) is deeply appreciated.

I have modified your title in order that your question is better refrenced. Do you agree. — Jean Marie, May 02 '17 at 14:41

score 6 · Accepted Answer · answered May 02 '17 at 13:27

6

Assume without loss of generality that $z_0=0$. We have \begin{align} q(z)&=q'(0)\,z+\frac{q''(0)}{2}\,z^2+O(z^3)\\ (q(z))^2&=(q'(0)^2)\,z^2+q'(0)\,q''(0)\,z^3+O(z^4)\\ \frac{1}{q(z)^2}&=\frac{1}{(q'(0)^2)\,z^2}\Biggl(\frac{1}{1+\dfrac{q''(0)}{q'(0)}\,z+O(z^2)}\Biggr)\\ &=\frac{1}{(q'(0)^2)\,z^2}\Bigl(1-\frac{q''(0)}{q'(0)}\,z+O(z^2)\Bigr). \end{align}

answered May 02 '17 at 13:27

Julián Aguirre

76,354

First question, why may we assume WLOG ? – some1fromhell May 02 '17 at 13:31
@some1: Because we can translate any given problem to an equivalent problem. (and, the equivalent problem even has the same answer, rather than just giving a result that we have to transform back to be the answer to the original problem) – May 02 '17 at 13:36
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Consider $f(z+z_0)$. $0\leftrightarrow z_0$ and $z\leftrightarrow z-z_0$. – Julián Aguirre May 02 '17 at 16:04
I see, thank you. Last question though, how did you get from 2nd last line to the last line ? negative binomial ? – some1fromhell May 03 '17 at 04:56
$$\frac{1}{1+z}=1-z+O(z^2)$$ – Julián Aguirre May 03 '17 at 12:04

score 2 · Answer 2 · answered May 02 '17 at 14:58

2

Hint: $f$ has a pole of order 2 at $z_0$ (why?), so the Laurent series of $f$ will be $$f(z) = \sum_{n=-2}^\infty a_n(z - z_0)^n,$$ while $$ q(z) = \sum_{n=1}^\infty b_n(z - z_0)^n,\qquad\text{with } b_n = \cdots $$ Use now that $$f(z)q(z)^2 = 1.$$

answered May 02 '17 at 14:58

Martín-Blas Pérez Pinilla

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Proof of the residue formula Res$(f(z);z_0) = -\frac{q''(z_0)}{(q'(z_0))^3}$ for a pole of order two

2 Answers2