# Find the Laurent series of $f(z)=\frac{1}{z(1-z)}$

I am having difficulties finding Laurent series of the above function, around these two domains: $$0<|z-1|<1$$ and $$|z-1|>1$$

The function $f(z)$ takes the form $\dfrac{1}{z}-\dfrac{1}{z-1}$.

And I know that I can find the Laurent series of the following general function around $z_0=0$ when my domain includes $z_0=0$:

$$\dfrac{1}{z-a}=\dfrac{1}{z}\cdot \dfrac{1}{1-\frac{a}{z}}=\sum_{n=1}^\infty\frac{a^{n-1}}{z^n}$$

But now I should evaluate the expansion around $z_0=1$. How can this be done?

Should I expect a Laurent series for the left partial fraction term?

An explanation about this would be very respected. It just that I didn't find a similar question when searching for it.

• Avoid using display styles in the title. :-) – Zain Patel Jul 1 '15 at 19:44

## 2 Answers

It is just a matter of writing it in a way that the series will converge. The strategy almost surely will be to use the geometric series.

Suppose $0 < |z-1|< 1$. We have: \begin{align}\frac{1}{z(1-z)} &= \frac{1}{z} - \frac{1}{z-1} \\ &= \frac{1}{1 + z - 1} - \frac{1}{z-1} \\ &= \frac{1}{1 -(-(z-1))} - \frac{1}{z-1} \\ &= \sum_{n \geq 0} (-1)^n(z-1)^n - \frac{1}{z-1} \\ &= -\frac{1}{z-1} + 1 -(z-1) +(z-1)^2 -(z-1)^3 + \cdots \end{align}, and $|z-1| < 1$ ensures convergence of my expansion above.

Now suppose $|z-1| > 1$. I want to use the geometric series again, but this only tells me that $1/|z-1| < 1$, so I'd better use geometric series for $1/(z-1)$. We have that $1/(z-1)$ is already good to go, so you only need to rewrite that $1/z$ term as something like $\frac{\rm stuff}{{\rm stuff} - \left(\frac{1}{z-1}\right)}$. Can you do it now?

Another way to look at it: using the geometric series formula, $a + ar + ar^2 + \dots = \dfrac{a}{1-r}$, we have $$\dfrac{1}{z-z^2} = \dfrac{\frac{1}{z}}{1 - z} = \dfrac{1}{z} + 1 + z + z^2 + \dots$$