2
$\begingroup$

I am trying to evaluate the integral $$\int \exp\left(-\frac{1}{z}\right)\sin\left(\frac{1}{z}\right)dz$$ in the deleted neighborhood $|z|=1$. This integral can be easily solved using the Cauchy integral formula, but this requires figuring out the the Laurent series since $z=0$ is an essential singularity. Could someone please show me how to write the Laurent series in order to find the $a_{-1}$ term. Thank you

$\endgroup$
1
  • $\begingroup$ writing $\sin (1/z) = (e^{i/z} - e^{-i/z})/(2i)$ and using expression of the exponential function. $\endgroup$
    – nguyen0610
    Apr 17, 2017 at 21:24

1 Answer 1

2
$\begingroup$

$\exp(-w) \sin(w)$ is an entire function. It has Maclaurin series $$ \exp(-w)\,\sin(w) = w - w^2 + \frac{1}{3}\,w^3 + \dots $$ valid in the whole complex plane. Therefore your function has a convergent Laurent series $$ \exp\left(-\frac{1}{z}\right)\sin\left(\frac{1}{z}\right) = \frac{1}{z} - \frac{1}{z^2} +\frac{1}{3z^3} +\dots $$ valid for all $z \ne 0$.

$\endgroup$
1
  • $\begingroup$ Thank you for the response, now I understand. $a_{-1}$=1 . $\endgroup$ Apr 18, 2017 at 12:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .