- #1
Arya Prasetya
- 8
- 3
Homework Statement
Find Laurent series of $$z^2sin(\frac{1}{1-z})$$ at $$0<\lvert z-1 \rvert<\infty$$
Homework Equations
sine series expansion.
The Attempt at a Solution
At first, it seems pretty elementary since you can set
[tex] w=\frac{1}{z-1} [/tex] and expand at infinity in z, which is 0 in w. Therefore, $$z^2sin(\frac{1}{z-1})=(\frac{1}{w^2}+\frac{2}{w}+1)sin(w)$$
$$=(\frac{1}{w^2}+\frac{2}{w}+1)\sum_{n=0}^\infty\frac{(-1)^nw^{2n+1}}{(2n+1)!}$$
$$=\sum_{n=0}^\infty\frac{(-1)^nw^{2n-1}}{(2n+1)!}+2\sum_{n=0}^\infty\frac{(-1)^nw^{2n}}{(2n+1)!}+\sum_{n=0}^\infty\frac{(-1)^nw^{2n+1}}{(2n+1)!}$$
But now, I am having troubles adjusting the sums so that the coefficients coincide into one series. The problem I have right now is the change of indices. I believe it is possible to do it for the 1st and 3rd term so that it becomes one. However, It might be impossible incorporating the term contributing to even powers of the w into one sum. Is it the case that Laurent series is able to be represented as 2 or more sums and have its coefficient on an even-odd case basis? Have I done something wrong?
Looking forward for the help and thanks in advance.