1. ## residue

I have an integral to evaluate via contour integration:
$({1 \over 1+cos^2\theta}) \ wrt d\theta$
Using the substitution $z =e^{i\theta}$I find the integral becomes,with $d\theta \ becoming \ {dz \over iz}$
${4z^2 \over z^4 + 6z^2 + 1}$ is that correct?
Also how would I evaluate this via the residue theorem?

2. That depends what you're integrating. For example, suppose:

$\int_0^{2\pi}\frac{1}{1+\cos^2(t)}dt$

and I let $z=e^{it}$, then I'd get:

$-i\mathop\oint\limits_{|z|=1}\frac{4z}{z^4+6z^2+1}d z=(-i) 2\pi i \sum\mathop\text{Res}\frac{4z}{z^4+6z^2+1}$

where the sum is over the residues of the enclosed poles.

3. I've written it differently:

${1\over i} \oint ({3\over2z}+{z\over4}+{1\over4z^3})^{-1}dz$

how can I pick the relevant term (the cofefficient of 1/z) to give the residue? The inverse is complicating matters....

4. I don't see why you'd want to write it that way. Just factor the denominator of the expression I got, get the enclosed zeros, bingo-bango.

5. Originally Posted by shawsend
I don't see why you'd want to write it that way. Just factor the denominator of the expression I got, get the enclosed zeros, bingo-bango.
Thanks.
So solve $z^4 + 6z^2 +1 =0$ to begin with...? But how would I?

6. Originally Posted by bigdoggy
Thanks.
So solve $z^4 + 6z^2 +1 =0$ to begin with...? But how would I?
Let $z^2=k$..."bingo-bango"!

7. Originally Posted by Drexel28
Let $z^2=k$..."bingo-bango"!
Hi, which gives $k=-3 \pm2\sqrt2 =z^2$
Am I doing what you're implying...!? But how can I get the poles here...?

8. Originally Posted by bigdoggy
Hi, which gives $k=-3 \pm2\sqrt2 =z^2$
Am I doing what you're implying...!? But how can I get the poles here...?
The poles are merely the zeros of the denominator. You know that $z^2=-3\pm \sqrt{2}\implies\text{Denominator equals zero}$ therefore $z=?\implies\text{Denominator equals zero}$?

9. Originally Posted by Drexel28
The poles are merely the zeros of the denominator. You know that $z^2=-3\pm \sqrt{2}\implies\text{Denominator equals zero}$ therefore $z=?\implies\text{Denominator equals zero}$?
I can't see it being $z= \pm\sqrt{-3\pm \sqrt{2}}$
I'm missing something here ...

10. Originally Posted by bigdoggy
I can't see it being $z= \pm\sqrt{-3\pm \sqrt{2}}$
I'm missing something here ...
You screwed up the quadratic equation.

- Wolfram|Alpha -Step 1

- Wolfram|Alpha -Step 2...validattion.

11. Thankyou.

So if the integrand is around |z|=1, the solution is $z=\pm i\sqrt{3 -2\sqrt2}$
I can't see how this would give a 'nice' (multiple of pi) answer (which I was told it should!)....

I find the residue = $1 \over \sqrt2$ correct??

12. The zeros are $z_0=i\sqrt{3+2\sqrt{2}},z_1=i\sqrt{3-2\sqrt{2}}, z_2=-i\sqrt{3+2\sqrt{2}},z_3=-i\sqrt{3-2\sqrt{2}}$. Now, which ones are in the unit circle? Isn't it the $3-2\sqrt{2}$ ones? So then we have:

$\int_0^{2\pi} \frac{1}{1+\cos^2(t)}dt=2\pi \mathop\text{Res}\limits_{z_1,z_3} \left\{\frac{4z}{z^4+6z^2+1}\right\}$.

I'll do one of the residues:

$r_1=\frac{4z}{(z-i\sqrt{3+2\sqrt{2}})(z+i\sqrt{3+2\sqrt{2}})(z+i\sq rt{3-2\sqrt{2}})}\biggr|_{z=i\sqrt{3-2\sqrt{2}}}=\frac{1}{2\sqrt{2}}$

Edit: Also, forgot to mention, if it's a simple pole and the derivative of the denominator is not zero there then:

$\mathop\text{Res}\limits_{z=z_0} \frac{f}{g}=\frac{f(z_0)}{g'(z_0)}$

May be easier to calculate these residues this way.