Find laurent expansion of $\frac{z-1}{(z-2)(z-3)}$ in annulus {$z:2<|z|<3$}.

Find the Laurent expansion of $\frac{z-1}{(z-2)(z-3)}$ in annulus {$z:2<|z|<3$}.

So far I have the following; I'm not 100% sure if it is right.

$\frac{z-1}{(z-2)(z-3)}$ = $\frac{2}{(z-3)}$-$\frac{1}{(z-2)}$

For $\frac{1}{(z-2)}$=$\frac{1}{z}$ $\frac{1}{1-(\frac{2}{z})}$=$\frac{1}{z}$$\sum_{k=1}^n\frac{2^k}{z^k}$=$\sum_{k=1}^n\frac{2^k}{z^{k+1}}$ for $|z|<1$.

I am having trouble with the other fraction. I have seen similar questions asked, but I cannot seem to get the information I need. Any input would be much appreciated!

Solution 1:

$$\frac{z-1}{(z-2)(z-3)}=\frac{2}{z-3}-\frac{1}{z-2}=\frac{2}{z-3}-\frac{1}{z}\frac{1}{1-\frac{2}{z}}=$$

$$=\frac{2}{z-3}-\frac{1}{z}\left[1+\frac{2}{z}+\left(\frac{2}{z}\right)^2+...\right]$$

Note that the first summand above is analytic in the given annulus, and the above expansion of $\,\dfrac{1}{z-2}\,$ applies whenever

$$\left|\frac{2}{z}\right|<1\Longleftrightarrow |z|>2$$

Solution 2:

Related problems: (I), (II). Use partial fraction to write your function as

$$\frac{z-1}{(z-2)(z-3)}= - \frac{1}{z-2} + \frac{2}{z-3} = -\frac{1}{z(1-\frac{2}{z})}-\frac{2}{3(1-\frac{z}{3})} $$

$$ = -\frac{1}{z}\sum_{k=0}^{\infty}\frac{2^k}{z^{k}}-\frac{2}{3}\sum_{k=0}^{\infty}\frac{z^k}{3^k} \,. $$

Now, the first series converges in $|z|>2$ and the second series converges in $|z|<3$ which implies the common region is $ 2< |z| < 3 $