I'm having a hard time finding the Laurent series for the following question :
$$f(z)=\sin\left(\frac{z+2}{z}\right)$$
for $U = \mathbb{C}^{*} = \left\{0 < |z| < \infty \right\}$
I tried using the following formula.
$$\sin(z+w) = \cos(w)\sin(z)+\cos(z)\sin(w)$$
I rewrite $f(z)=\sin(1)\cos\left(\frac{2}{z}\right)+\sin(\frac{2}{z})\cos(1)$,
but I'm stuck here. Any help would be a lot appreciated. Thanks in advance!
The Taylor series for $\sin$ and $\cos$ on the whole plane are $\sin(z)=z-\frac{1}{3!}z^3+\frac{1}{5!}z^5-\dots$ and $\cos z=1-\frac{1}{2!}z^2+\frac{1}{4!}z^4-\dots$ (we can proof this by (i) calculating the $n$-th derivative or (ii) using the series $\exp=1+x+\frac{x^2}{2!}+\dots$ and the identity $\exp iz=\cos z+i\sin z$).
Plug into your formula, we have: $$f(z)=\sin(1)(1-\frac{2^2}{z^22!}+\frac{2^4}{z^44!}-\dots)+\cos 1(\frac{2}{z}-\frac{2^3}{z^33!}+\frac{2^5}{z^55!}-\dots)$$
By a rearragement, we obtain the Laurent series.