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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 $z$?

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

$$\frac1z\sin2x=\frac1z\sum_{n=1}^\infty(-1)^{n-1}\frac{(2z)^{2n-1}}{(2n-1)!}=\sum_{n=1}^\infty(-1)^{n-1}\frac{2^{2n-1}z^{2n-2}}{(2n-1)!}$$

and we get in fact a power series, as expected.

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  • $\begingroup$ I actually got $$\sum \limits_{n=0}^{\infty} (-1)^n \frac {2^{2n+1}z^{2n}}{(2n+1)!}$$ this is the same thing right? $\endgroup$ – snowman May 26 '15 at 15:25
  • $\begingroup$ @snowman Yes, indeed: same thing. $\endgroup$ – Timbuc May 26 '15 at 15:29

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