Laurent Expansion of $\frac{1}{f(z)^2}$, for $f$ a holomorphic function on a nbd of $a \in C$ with $f(a) = 0$ $f'(a) \ne 0$ and $f''(a) = 0$.

Question

First, we are asked:

For $f$ a holomorphic function on a nbd of $a \in C$ with $f(a) = 0$ $f'(a) \ne 0$, find the residue of $\frac{1}{f(z)}$. I used $f'(a) \ne 0$ to conclude the $f$ is not constant, and thus $a$ is an isolated zero of degree $n \in N$ of $f$. Therefore, $a$ is a pole of degree $n$ of $\frac{1}{f}$. Thus the residue is the standard formula for a pole of degree $n$. Is this correct so far?

Next we are asked:

For $f$ as above, also assume that $f''(a) = 0$. Then find the principle part of the Laurent Expansion of $\frac{1}{f(z)^2}$.

I understand that the principle part is the sum of negative-index entries in the Laurent series, but how do I find it for $\frac{1}{f(z)^2}$? I tried to find some connection between the Cauchy formula for $f''(a)$ and the formula for Laurent coefficients, but cannot see how these relate I assume the answer lies elsewhere..?

Answer 1

As $f'(a)\neq 0$, you know the order of the pole of $\frac1{f(z)}$ at $a$ is $1$.

You have $f(z)=f'(a)(z-a)+f''(a)(z-a)^2/2+\dots=(z-a)f'(a)\left(1+\frac{f''(a)}{2f'(a)}(z-a)+\dots\right)$.

Plugging into the geometric series yields $$\frac{1}{f(z)} =\frac{1}{f'(a) (z-a)}\left( 1-\frac{f''(a)}{2f'(a)} (z-a)+\dots\right) $$ so the residue of $\frac1{f(z)}$ at $z=a$ is $\frac{1}{f'(a)}$.

If also $f''(a)=0$ the Taylor series starts with $f(z)=f'(a)(z-a)+\frac{f'''(a)}{6}(z-a)^3+\dots=f'(a)(z-a)\left( 1+\frac{f'''(a)}{6f'(a)}(z-a)^2+\dots\right)$ so the derivative of the geometric series tells you $$ \frac{1}{(f(z))^2}=\frac{1}{(f'(a))^2(z-a)^2}\left(1-2\frac{f'''(a)}{6f'(a)}(z-a)^2+\dots \right) $$ so the principal part of the Laurent series is just $\frac{1}{(f'(a))^2(z-a)^2}$.

By "plugging into the geometric series" I mean here that we use that if $f(a)=0$, the Taylor series near $z=a$ of $\frac{1}{1-f(z)} $ can be computed as $$\frac{1}{1-f(z)}=1+f(z)+(f(z))^2+(f(z))^3+\dots $$ by using that $f(z)=b_1(z-a)+b_2(z-a)^2+b_3(z-a)^2+\dots$, you obtain taking powers of the series and combining that $$\frac{1}{1-f(z)}=1+(b_1(z-a)+b_2(z-a)^2+b_3(z-a)^2+\dots)+(b_1^2(z-a)^2+2b_1b_2(z-a)^3+\dots)+(b_1^3(z-a)^3+3b1^2b_2(z-a)^4+\dots= b_1(z-a)+(b_2+b_1^2)(z-a)^2+(b_3+2b_1b_2+b_1)^3(z-a)^3+\dots. $$ Using Faa di Bruno's formula, the result of this must be the Taylor expansion of $\frac{1}{1-f(z)}$, hence a convergent series.

Similarly you can use also $$\frac1{(1-f(z))^2}=1+2f(z)+3(f(z))^2+4(f(z))^3+\dots. $$