Laurent Series expansion of $f(z)=(z-1) \sin{1\over z}$

Question

I need to find the Laurent series expansion of the function $f(z)=(z-1) \sin{\dfrac 1 z}$ in $A= \{ z \in \Bbb C : 0<|z|<\infty \}$.

Any help would be appreciated!

Answer 1

It is known that $$\sin w = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \cdot w^{2n+1}.$$ Substitution $w = \frac{1}{z}$ yields $$\sin \frac{1}{z} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \cdot \frac{1}{z^{2n+1}}$$ so $$(z-1) \sin \frac{1}{z} = z \cdot \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \cdot \frac{1}{z^{2n+1}} - \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \cdot \frac{1}{z^{2n+1}} \\[1ex] = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \cdot \frac{1}{z^{2n}} - \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \cdot \frac{1}{z^{2n+1}}.$$