Laurent series of z^2sin(1/(z-1))

[Arya Prasetya]

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
w=\frac{1}{z-1} 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.

 

[fresh_42]

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
w=\frac{1}{z-1} 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.

It often helps to write down a couple of terms, say from $w^{-1}$ to $w^3$. Another possibility is to introduce new counters: $m=2n-1$ in the first series, $m=2n$ in the second, and $m=2n+1$ in the third. The second has only even powers, and the first and third have only odd powers, so they can be added. You'll get an extra term for $w^{-1}$ which occurs only once. In the end you'll get something like
$$
a_{-1}w^{-1} + \sum_{m \;even}a_mw^m + \sum_{m \;odd}a_mw^m
$$
which results in $\sum_{m \geq -1}a_mw^m$ with $a_m = \begin{cases} \ldots & \text{if } m = -1 \\ \ldots & \text{if } m \geq 0 \text{ and } m \text{ odd } \\ \ldots & \text{if } m \geq 0 \text{ and } m \text{ even } \end{cases}$

 

[Arya Prasetya]

Ahh ok, got it got it, that's what I thought. Thanks for the insight! :)