I'm having a hard time finding the Laurent series for the following question :


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!

  • $\begingroup$ Thanks for your edits, I didn't know there was \sin et \cos function, I'm new in latex. I wanna know if I can write $\sin 2z^{-1} =2 [ z^{-1} - \frac{z^{-3}}{3!} + \frac{z^{-5}}{5!} - \frac{z^{-7}}{7!} + \ldots ] $ $\endgroup$ – JoshuaK Apr 17 at 1:19
  • 1
    $\begingroup$ That's almost right, except the $2$ needs to be in every term. $\endgroup$ – Ninad Munshi Apr 17 at 1:25
  • $\begingroup$ Yes I wrote $2\sin(z^{-1})$ my bad thanks !! $\endgroup$ – JoshuaK Apr 17 at 3:14

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.

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