# Homework Help: Contour integral of e^(-1/z) around a unit circle?

1. Jun 9, 2013

### CrimsonFlash

1. The problem statement, all variables and given/known data
What is the integral of e-1/z around a unit circle centered at z = 0?

2. Relevant equations
-
3. The attempt at a solution
The Laurent expansion of this function gives : 1 - 1/z + 1/(2 z^2) - 1/(3! z^3) + . . . . .
The residue of the pole inside is -1.
So the integral should be = -2πi

Is this right? Also, aren't there infinitely many poles or infinitely high order of the pole at z = 0?

2. Jun 9, 2013

### Staff: Mentor

So what? The residue is enough to find the answer, and it looks right to me.

3. Jun 9, 2013

### Mute

Remember that the contour integral

$$\oint_{|z| = R} \frac{dz}{z^n}$$
around a circle of radius R centered at the origin is zero for any $n \neq 1$. You can show this explicitly by the change of variables $z = \exp(i\phi)$. (You can also show the integral is $2\pi i$ for n = 1 this way).

So, as long as you can exchange the sum with the integral (which you can in this case), all terms but the residue term will integrate to zero.

4. Jun 9, 2013

### BruceW

yeah, you got it right. And yes, it is a pole of infinitely high order (an essential singularity). But this is not a problem. If there were infinitely many poles, then there would be a problem. But you can see for yourself that for this case, the singularity is isolated. Imagine removing the point z=0, then are the other points around it holomorphic?

5. Jun 9, 2013

### jackmell

No. There is not even one pole. Poles are finite-ordered singular points. Rather, the singular point at the origin is an essential singularity with infinite order.

6. Jun 9, 2013

### CrimsonFlash

Ok, thanks everyone. Also, can you help me with the following:
f(z) = sqrt(z)/(1+z^2) = sqrt(z)/[(i+z)(z-i)]
The poles of this are at i and - i. The residues I calculate are:
For i := sqrt(i)/(2i) = exp(πi/4)/(2i)
For -i := sqrt(-i)/(-2i) = exp(-πi/4)/(-2i)

Bur WolframAlpha says otherwise. The i shouldn't be there in the denominator. But I can't see how to get rid of them...

7. Jun 9, 2013

### Staff: Mentor

1/i = -i, does that help?
And you can evaluate exp(πi/4) if you like.

8. Jun 9, 2013

### CrimsonFlash

Never mind, I got what I was looking for.
Thanks anyway.