Residue of $f(z) = (z-z_1)(z-z_2)(z-z_3)...(z-z_n)$ at $z_1$

Question

I found a solution to finding the residue of $1/f(z)$ when $f(z) = (z-z_1)(z-z_2)(z-z_3)...(z-z_n)$ at $z_1$ as $[(\frac{1}{z_1-z_2})(\frac{1}{z_1-z_3})...(\frac{1}{z_1-z_n})]$

Was this found using the Laurent expansion or some other formula? I don't understand how they were able to find this residue! Thanks!

Answer 1

When the limit $\require{cancel}\lim_{z\to a}(z-a)f(z)$ exists, then the residue is equal to that limit. So\begin{align}\operatorname{res}_{z=z_1}\frac1{(z-z_1)(z-z_2)\ldots(z-z_n)}&=\lim_{z\to z_1}\frac{\cancel{z-z_1}}{\cancel{(z-z_1)}(z-z_2)\ldots(z-z_n)}\\&=\frac1{(z_1-z_2)\cdots(z_1-z_n)}.\end{align}

Answer 2

For a simple pole at $z = c$ of a meromorphic function $f(z)$, one may always calculate residue by the formula $$ \operatorname{Res}(f,c) = \lim_{z\to c}f(z)(z-c) $$ Direct computation of the limit in your case yields the result.

Answer 3

Formally write $$ \frac{1}{f(z)}=\frac{a_{-1}}{z-z_{1}}+a_0+a_1(z-z_1)+a_2(z-z_2)^2+\dotsb $$ multiply both sides by $z-z_1$ and let $z\to z_1$ to get $$ a_{-1}=\lim_{z\to z_1}\frac{z-z_1}{f(z)}=\frac{1}{(z_1-z_2)(z_1-z_3)\dotsb(z_1-z_n)} $$

Answer 4

Hint: Any time you have a function of the form

$$\frac{1}{z-z_1}\cdot g(z),$$

where $g$ is analytic in some disc $D(z_1,r),$ then the residue of that function at $z_1$ equals $g(z_1).$