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

[CrimsonFlash]

Homework Statement

What is the integral of e^-1/z around a unit circle centered at z = 0?

Homework Equations

-

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?

 

[mfb]

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

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

 

[Mute]

Homework Statement

What is the integral of e^-1/z around a unit circle centered at z = 0?

Homework Equations

-

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?

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.

 

[BruceW]

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

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?

 

[jackmell]

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

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.

 

[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...

 

[mfb]

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

 

[CrimsonFlash]

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