# 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. :-) Commented Jul 1, 2015 at 19:44

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 \\&=\sum_{n=-1}^\infty (-1)^n(z-1)^n\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?
• Why for $|z-1|<1$ you don't expand $\frac{1}{z-1}$? Is it like $\frac{1}{z}$ around zero, where it is it's own Taylor expansion. Commented Feb 10, 2020 at 10:47
• Is that is because $\frac{a_{−1}}{z−1}$ with $a_{−1}=1$ (in this case) is a principle part with one term only for the region $0<|z-1|<1$. In the second case, however, for $|z-1|>1$? Shouldn't both $\frac{1}{z}$ and $\frac{1}{z-1}$ be expanded? Commented Feb 10, 2020 at 14:18
Another way to look at it: using the geometric series formula, $a + ar + ar^2 + \dots = \dfrac{a}{1-r}$, we have $$\dfrac{1}{z-z^2} = \dfrac{\frac{1}{z}}{1 - z} = \dfrac{1}{z} + 1 + z + z^2 + \dots$$