Find the Laurent series expansion of $\frac{z}{(1-z)^2}$

Question

I want to find the $g(z) = \frac{z}{(1-z)^2}$ for $z \in \mathbb{C}-\{1\}$. My approach has been to decompose the fraction, i.e. $$\frac{z}{(1-z)^2} = \frac{A}{1-z} + \frac{B}{(1-z)^2} \implies A(1-z) + B = z \iff A = -1 , B = 1 \implies g(z) = - \frac{1}{1-z} + \frac{1}{(1-z)^2}.$$

Now I want to rewrite the sum in terms of power series. The first term becomes $$-\frac{1}{1-z} = - \sum_{k=0}^{\infty} z^k,$$

and the second one I do not know how to get.. This is my problem and I'd like some help. If I restrict the domain to be $|z| > 1$ the second term becomes $$\frac{1}{(1-z)^2} = \sum_{k=-\infty}^{-2}(k+1)z^k.$$

If $|z| < 1$, then we have $\frac{1}{(1-z)^2} = (\sum_{k=0}^{\infty} z^k)^2$, and I do not know how to continue from here. I found an article on MSE, but didn't know how to apply it to my problem and I also don't think it should be that 'complicated' in my exercise.

All help is appreciated. Thanks!

Answer 1

Hint Notice that we can also rewrite $g$ with some simple algebraic manipulation: $$\frac{z}{(1 - z)^2} = -\frac{1 - z}{(1 - z)^2} + \frac{1}{(1 - z)^2} = \frac{1}{(1 - z)^2} - \frac{1}{1 - z} .$$

To write a series for this expression centered at $\{z = 0\}$, we can use the observation $$\frac{1}{(1 - z)^2} = \frac{d}{dz} \left(\frac{1}{1 - z} \right). $$

Formally (and analytically on $\{|z| < 1\}$), we thus have $$\frac{1}{(1 - z)^2} = \frac{d}{dz} \left(\sum_{k = 0}^\infty z^k \right) = \sum_{k = 1}^\infty k z^{k - 1} = \sum_{l = 0}^\infty (l + 1) z^l .$$ Alternatively, reindexing $$\left(\sum_{k = 0}^\infty z^k\right)^2 = \sum_{k = 0}^\infty z^k \sum_{k' = 0}^\infty z^{k'} = \sum_{k = 0}^\infty \sum_{k' = 0}^\infty z^{k + k'}$$ by total degree gives $$\sum_{l = 0}^\infty \left(\sum_{k_1 + k_2 = l} 1\right) z^l = \sum_{l = 0}^\infty (l + 1) z^l .$$