Question about how to calculate a certain residue

Question

I want to calculate the residue at $z=i$ for $f(z)=\frac{1}{(z^2+1)(z^2+9)(z^4+4)}$.

I've calculated the residue correctly by using the formula $Res(f;i)=\lim_{z\to i}{(z-i)f(z)}$.

I also know that it is possible to use the formula: $Res(f;i)=\frac{P(i)}{Q'(i)}=\frac{1}{D((z^2+1)(z^2+9)(z^4+4))|_{z=i}}$

But in the solution to the problem the following calculation is done:

$Res(f;i)=\frac{1}{(z^2+9)(z^4+4)D(z^2+1)|_{z=i}}$

Greatful if anyone can show me how this formula follows from the other two formulas, or give me another explanation to why it works.

What is $D$? Is it a function or a number? – Nikita Evseev May 14 '15 at 07:57 — Nikita Evseev, May 14 '15 at 07:57
It's the differentiation operator. – Jarvi79 May 14 '15 at 08:58 — Jarvi79, May 14 '15 at 08:58

score 2 · Accepted Answer · answered May 14 '15 at 09:09

2

They use $P(z) = \frac{1}{(z^2+9)(z^4+4)}$ and $Q(z) = z^2+1$.

answered May 14 '15 at 09:09

Nikita Evseev

1 Answers1