# Solving integral in complex analysis

1. Oct 27, 2013

### FreySmint

Hi! I'm new here, been a fan of this site for years, but only now I felt the need of registering.

1. The problem statement, all variables and given/known data

Use the Residue Theorem to solve the integral:

∫[(cos(2t)) / (5-4*cos(t)) ] dt from t=0 to t=2pi

2. The attempt at a solution

I did a variable change z=eit. With that:

(1/(2i))*∫[ (z2 + z-2) / (-2z2 +5z -2)]

Only taking the integral, I came up with:

∫[ (z4 +1) / (z2 * (z-1/2) * (z-2))]

Which has 3 singularaties: 1/2 , 2 and 0
Only 2 is not in the path considered.

The residue in 1/2 is -17/6 and in 0 is 5/2 as they are poles of order 1 and 2 respectively.

And with that, I have:

1/(2i) * 2*pi*i * (-17/6 + 5/2) = - Pi/3

Which is wrong, as the answer is actually Pi/6

Where did I go wrong?

Last edited: Oct 27, 2013
2. Oct 27, 2013

### jackmell

Re-do the residue calculations

3. Oct 27, 2013

### Dick

And notice that (z-1/2)*(z-2) is not equal to (-2z^2+5z-2).

4. Oct 28, 2013

### jackmell

ouch. Would have failed that test. Thanks for pointing that out Dick. Hope the op didn't spend a lot of time re-doing the residue calculations with the wrong factor.

5. Oct 28, 2013

### vanhees71

Note that to get your original integral you have to integrate
$$f(z)=\frac{1}{\mathrm{i} z} \cdot \frac{z+z^{-2}}{2(5-2(z+1/z))}=\frac{\mathrm{i}(1+z^4)}{z^2(2z^2-5z+2)}$$
along the unit circle $z=\exp(\mathrm{i} t)$ with $t \in \{0,2 \pi\}$.

Now you can use the residue theorem, which gives you immediately $\pi/6$. To be honest, I've been lazy and did this with help of Mathematica :-).

6. Oct 29, 2013

### FreySmint

Thanks everyone! I feel really stupid right now, my mistake was quite dumb... I had forgotten that when you use quadratic formula, if the quadratic coefficient (a) is not 1, you'll have a*(x-z1)(x-z2). I often forget the most basic stuff.
Thank you!