Find the Laurent series in given annulus:
I have function in annulus $1<|z-1|<2$, $${1 \over {z(z-3)^2}}$$
After splitting function into partial fractions, I got the first two sums, but stacked in last one
$${1\over 3(z-3)^2}$$ What should i do for that?
Any help or advice are welcome!
Note that$$\frac1{3(z-3)^2}=\frac13(z-3)^{-2}=\frac13\left(\frac1{3-z}\right)'.$$So, compute the Laurent series of $\dfrac1{3-z}$ and differentiate it termwise.