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 $z$?

  • $\begingroup$ $f(z)=\frac{\sin(2z)}{z}$ 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

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:


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.