Laurent Series of $\mathrm e^z/((z - 1)(z - 2))$ for $1 < | z |<2$. I am working at my homework and try to calculate the Laurent series of the following function
$$ f(z) = \frac{\mathrm e^z}{(z - 2)(z-1)}$$
for $1 < \lvert z \rvert < 2$. So I have the annulus with radius $1$ with center $0$.
Now I know that
$$\frac{\mathrm e^z}{(z - 2)(z-1)} = \frac{\mathrm e^z}{(z - 2)} + \frac{\mathrm e^z}{(1 - z)}.$$
Moreover, $\mathrm e^z = \sum_{n = 0}^\infty \frac{z^n}{n!}$,
$$ \frac{\mathrm e^z}{(z - 2)} = \frac{\mathrm e^z}{z(1 - 2/z)}$$ and I know that $\frac{1}{(z - 2)} = \frac{1}{z}\sum_{n = 0}^\infty(2/z)^n$ for $\lvert z \rvert < 2$. So I can write the first term as a power series. But I don't know how to develop the second term. Furthermore, I am also not sure how to get the multipliers $a_k$ such that $\frac{\mathrm e^z}{(z - 2)} = \sum_{k = -1}^\infty a_k z^k$. What am I missing?

The expansion for $|z| \lt 2$
$$
\frac{1}{z-2} = \frac{-1/2}{1-z/2}\\
= -\frac{1}{2} \sum_{n=0}^\infty \bigg( \frac{z}{2} \bigg)^n
$$
The expansion for $|z| \gt 1$ of $\frac{1}{z-1}$ follows what you have for $z-2$ as expanding in $\frac{1}{z}$
$$
\frac{1}{z-1} = \frac{1}{z} \frac{1}{1-1/z}\\
= \frac{1}{z} \sum_{n=0}^\infty \bigg( \frac{1}{z} \bigg)^n
$$
Combine these along with the $e^z$ series which also holds on this annulus.
