# Laurent series in domain $|z|>0$

Find Laurent series, in powers of $z$, of $$f(z)=\frac{\sin(2z)}{z}$$ valid in the region $|z|>0$.

The singularity is $0$ but $0$ isn't inside the region of the domain so what do you exactly expand?

Do you just expand $\sin(2z)$ and then divide it by $z$?

• $f(z)=\frac{\sin(2z)}{z}$ has a removable singularity in zero. Once you remove it you get an entire function: $$2\,\text{sinc}(2z)=\sum_{n\geq 0}\frac{4^{n+1}(-1)^n}{(2n+1)!}z^{2n}.$$ – Jack D'Aurizio May 26 '15 at 15:08
• what is sinc? ??? – snowman May 26 '15 at 15:13
• en.wikipedia.org/wiki/Sinc_function – Jack D'Aurizio May 26 '15 at 15:18

The function's analytic in your region (and almost analytic at $\;z=0\;$ ...), so using the power series for $\;\sin z\;$ which has infinite convergence radius:
$$\frac1z\sin2x=\frac1z\sum_{n=1}^\infty(-1)^{n-1}\frac{(2z)^{2n-1}}{(2n-1)!}=\sum_{n=1}^\infty(-1)^{n-1}\frac{2^{2n-1}z^{2n-2}}{(2n-1)!}$$
• I actually got $$\sum \limits_{n=0}^{\infty} (-1)^n \frac {2^{2n+1}z^{2n}}{(2n+1)!}$$ this is the same thing right? – snowman May 26 '15 at 15:25