Finding the laurent expansion of $\frac{z}{(z-1)(z-2)}$ I want to  find the Laurent series for $\frac{z}{(z-1)(z-2)}$ in the region $1 < |z| < 2$. This implies that $\frac{1}{|z|} < 1$, so noticing that $(z-1) = z(1 - \frac 1 z)$ I can rewrite the desired function as
$$\frac{z}{z(1- \frac 1 z)(z-2)} = \frac{1}{z-2} \cdot \frac{1}{1 - \frac 1 z}.$$
Now using the definition of the geometric series I rewrite it as
$$\sum_{k=0}^{\infty} \frac{1}{z^k(z-2)}$$
Is this the right Laurent series, and if not, where did I go wrong? Please note I am trying to understand where I made a mistake, not simply finding any solution.
I've read a similar question at Finding the Laurent series of $f(z)=1/((z-1)(z-2))$ and one answer uses the fact that $\frac{|z|}2 < 1$, but I am not sure if my method is also valid.


Is this the right Laurent series, and if not, where did I go wrong?

The condition $1<|z|<2$ itself is sufficient to tell that each term of the Laurent series expansion must contain either negative or non-negative power of $z$ because the annular region $r<|z-z_0|<R$ between two concentric circles has center $z_0$ ; which is also the center of series. In your case, it is $0$. The correct expansion will be obtained by using partial fractions as displayed in other answers.
