how to expand laurent series of this function

Question

expand $f(z)=\frac{z^2}{(z+1)(z^2-1)}$ in regions:
a) $0<|z-1|<2$ and b) a) $2<|z-1|$ I need help with expanding this laurent series especially for region b. Thank you.

Answer 1

This is what I've tried and I hope it will help you:

b)

First rewrite $f$ as $$f(z)=\frac{z^2}{(z+1)^2(z-1)}=\frac{1}{z-1}g(z)$$ where $$g(z)=\left(\frac{z}{z+1}\right)^2= \left(1-\frac{1}{z+1}\right)^2=1-\frac{2}{z+1}+\frac{1}{(z+1)^2}$$ The domain $2<|z-1|$ can be seen as $|\frac{2}{z-1}|<1$ for which we cand find the expansion using the geometric series ( I recall that if $|w|<1$ then $\frac{1}{1-w}$ = $\sum_{n=0}^\infty w^n$ ).

$$\frac{2}{z+1}=\frac{2}{z-1+2}=\frac{2}{z-1}\cdot\frac{1}{1+\frac{2}{z-1}}$$ In this case, $w=-\frac{2}{z-1}$, so $\frac{1}{1+\frac{2}{z-1}}$ will be $\sum_{n=0}^\infty (-1)^n\frac{2^n}{(z-1)^n}$.

$$g(z)=1-\frac{2}{z-1}\sum_{n=0}^\infty(-1)^n\frac{2^n}{(z-1)^n}+\frac{1}{(z+1)^2}$$

To find the Laurent expansion of $\frac{1}{(z+1)^2}$ I suggest you to use the differentiation of the power series. So if $$\frac{1}{z+1}=\frac{1}{z-1}\sum_{n=0}^\infty (-1)^n\frac{2^n}{(z-1)^n}=\sum_{n=0}^\infty (-1)^n\frac{2^n}{(z-1)^{n+1}}$$ then $$-\frac{1}{(z+1)^2}=\sum_{n=1}^\infty -(-1)^n\frac{2^n(n+1)}{(z-1)^{n+2}}$$

Take care that now $n$ starts from $1$. After multiplying both terms by -1, you finally get that: $$g(z)=1-2\sum_{n=0}^\infty(-1)^{n}\frac{2^n}{(z-1)^{n+1}}+\sum_{n=1}^\infty (-1)^n\frac{2^n(n+1)}{(z-1)^{n+2}}$$

$$f(z)=\frac{1}{z-1}\left(1-2\sum_{n=0}^\infty(-1)^{n}\frac{2^n}{(z-1)^{n+1}}+\sum_{n=1}^\infty (-1)^n\frac{2^n(n+1)}{(z-1)^{n+2}}\right)$$

Answer 2

I'll do part a) since b) is already taken care of.

Rewrite the function as $f(z) = \frac{z^2}{(z+1)^2(z-1)}$

We can see that $f(z)$ is not analytic at the center of the disk $0<|z-1|<2$ so it is required to decompose it into partial fractions

For the first coefficient: $A = \lim_{z \to -1} ~(z+1)^2~f(z) = \lim_{z \to -1}~\frac{z^2}{z-1} = -\frac{1}{2}~~~~~~$.... etc

So, the function is described in partial fractions as: $$f(z) = \frac{1/4}{z-1} + \frac{3/4}{z+1} - \frac{1/2}{(z+1)^2}$$

Note that the first term is not analytic in the domain so its series will not converge on the disk, further, we want the series to be centered at the disk center $z=1$, thus we pursue a series in powers of $(z-1)$ and rearrange as follows: $$f(z) = \frac{1/4}{z-1} + \frac{3/8}{1-[-\frac{z-1}{2}]} + \frac{d}{dz}~\frac{1/4}{1-[-\frac{z-1}{2}]}$$ $$\\$$ $$==>~~~~f(z) = \frac{1/4}{z-1} + 3\sum_{k=0}^{\infty} \frac{(-1)^k}{2^{k+3}}~(z-1)^k + \sum_{k=1}^{\infty} \frac{(-1)^k}{2^{k+2}}~k~(z-1)^{k-1} $$ $$\\$$ $$~~~~~~~~~~~~~~= \frac{1}{2^2}(z-1)^{-1} + \frac{1}{2^2} - \frac{1}{2^4}(z-1) + 0~(z-1)^2 - \frac{1}{2^6}(z-1)^3 - ~...$$