# 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!

• Notice your expansion is valid for $\left|\frac{2}{z}\right| <1$, which is the exterior of the circle $|z|=2$. For the other term just notice $(z-3)^{-1} = -3^{-1}(1-\frac{z}{3})^{-1}$ and expand as a geometric series again. – Jose27 Nov 4 '12 at 0:28
• ah yes thank you, I forgot to add that in. thank you for the help, I just couldn't seem to see it – katherinebarry Nov 4 '12 at 0:47

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

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

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

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

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 $2< |z| < 3$