# Integrate over complex contour

1. Sep 23, 2011

### beefcake24

Evaluate the integral over the contour C when:

f(z) = 1/z and C = {z(t) = sin(t) + icos(t) | 0 <= t <= 2*pi}

I know f(z) = 1/r*e^(-it) = 1/r(cos(t) + isin(t)). But, when I try to take the contour integral by integrating f[z(t)]*z'(t), I get really messy formulas ((1/r*cos(sin(t)) + i*1/r*sin(cos(t)))*(cos(t) + -i*sin(t)), which makes me think I'm missing something.

Can anyone help me out on this? I have a midterm tomorrow and this was one of the practice questions.

2. Sep 23, 2011

### lanedance

what does the contour look like? do you think you could use the residue theorem?

3. Sep 23, 2011

### beefcake24

I don't know, we haven't covered the residue theorem yet, and we do not need to know the residue theorem for our midterm, so I think we should be able to solve this without it.

4. Sep 23, 2011

### beefcake24

The contour is just the unit circle, but starting at (0,i) when t=0 instead of (0,0) and traversed clockwise. Does this mean I can just take the negative of the integral around the unit circle? I know traversing a path in the opposite direction just changes the sign of the integral, and I don't think having a different starting point should affect the value, because you are traversing the same path. Can someone confirm this, or if I'm wrong, point me in the right direction?

5. Sep 23, 2011

### HallsofIvy

Staff Emeritus
Yes, so it would be better to write z in polar form. $z= cos(t)+ i sin(t)= e^{it}$
What are 1/z and dz?

That is pretty close to trivial to integrate.

Just as in real integration, integrating over the same path in the opposite direction (you would be integrating from $t= 2\pi$ to 0 rather than vice-versa, changes the sign.

If you want to do it more "directly", remember that $sin(z)= (e^{iz}- e^{-iz})/2i= -i(e^{iz}- e^{-iz})/2$ and $cos(x)= (e^{iz}+ e^{-iz})/2$ so that $sin(z)+ icos(z)= -i(e^{iz}- e^{-iz})/2+ i(e^{iz}+ e^{-iz})/2= e^{-iz}$. Again, just the previous formula with -iz rather than iz so that $dz= -ie^{-iz}dz$, and now you integrate from 0 to $2\pi$.

Last edited: Sep 23, 2011
6. Sep 23, 2011

### beefcake24

Perfect, that was just what I needed to finish the problem. Thanks!