2
$\begingroup$

I have a function that diverges like $e^{1/z}$ in the origin, and I'm integrating the function a closed contour in the complex plane. The point $z=0$ lies inside the contour, so if I integrate by residues the singularity at $z=0$ (a pole?) must be taken into account. How would you do that?

$\endgroup$
3
$\begingroup$

Hint. It's neither a pole nor a removable singularity. Write the Laurent series expansion of $e^{1/z}.$ This is a classic example where a function has an essential singularity (at $z=0.$)

$\endgroup$
  • $\begingroup$ If I write the Laurent expansion (as in math.stackexchange.com/questions/231133/…), it becomes a sum of infinte poles. Is that how you would compute the residue? $\endgroup$ – Jennifer Dec 29 '16 at 7:48
  • 1
    $\begingroup$ The residue would be 1. Because the contour integral around $z^{-n}$, for $n>1$ ($n$ being an integer), is zero, the residue only takes into account the $z^{-1}$ term of the Laurent series. $\endgroup$ – user361424 Dec 29 '16 at 7:50
4
$\begingroup$

$$e^z=1+\frac{z}{1!}+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdot\cdot\cdot, -\infty<z<\infty$$

This may also be written in the following manner.

$$e^z=\sum_{j=0}^\infty \frac{{z}^j}{j!}$$

Now, for $f(z)=e^{\frac{1}{z}}$ (which happens to have a singularity at $z=0$) we can write:

$$e^{\frac{1}{z}}=\sum_{j=0}^\infty \frac{(\frac{1}{z})^j}{j!}=\sum_{j=0}^\infty \frac{1^j}{j!\cdot z^j}$$

We know the residue would be the coefficient of the term containing the $-1^{th}$ power of $(z-z_o)$.

Coefficent of $j^{th}$ term would be:

$$\frac{1^j}{j!}$$

Clearly, the term we're looking for can be obtained at $j=1$.

So,

$$Res(e^{\frac{1}{z}})=1$$

$\endgroup$
  • 4
    $\begingroup$ Learn to use TeX in the site and write down an answer, please. You can find a tutorial in the Meta site. $\endgroup$ – Pedro Tamaroff May 13 '17 at 9:38

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.