Evaluate Definite Integral with Complex Analysis

[KleZMeR]

Homework Statement

$I_1 = \int_0^{2\pi} \frac{sin\theta}{3+2cos\theta} d\theta$

Homework Equations

Using identities to change from cos, sin, to variables of z, I get:

$2iz^2 + 6iz + 2i$ in my denominator

The Attempt at a Solution

Looking for a singularity, will I use a quadratic for this denominator to find my two zeros and then use my positive result to calculate the residue?[/B]

 

[KleZMeR]

Also, is this a pole of second order? Yes, I think it is. But what this means is that I will have to take a derivative of this fraction with polynomials? I have tried to simplify it into products but to no avail.

 

[vela]

You forgot to write $d\theta$ in terms of $dz$.

Yes, you need to use the quadratic equation to factor the denominator. All of the poles in this problem are simple poles.

 

[KleZMeR]

Thanks Vela, I had that written but didn't include that. So, I found my residue for my singularity at $z_0 = 0$
which was $-\frac{1}{2}$

But when I find my residue for the $z_0 = \frac{\sqrt5+3}{2}$ (the value I got from my quadratic)
I got a residue = 0.
Quadratic being $(z^2)+3z+1$

What do you think? I see from some examples I've done that this IS possible, but this is homework so I'd like to know what you think?

 

[vela]

Show your calculation of the residue. It shouldn't be 0. The integral, however, turns out to be equal to 0. You can see this by changing the interval of integration to $-\pi$ to $\pi$ and noting the integrand is an odd function.

 

[KleZMeR]

Well for starters I am not getting my ($z-z_0$) to cancel out, should I be using exponents instead of my z identities for the trig functions? The only reason I was able to get my residue at 0 was because $\frac{z-0}{z}$ occurred.

 

[vela]

From what you just said, it's not clear at all to me what you're doing to evaluate the residue.

 

[KleZMeR]

Using $cos\theta = \frac{1}{2}(z+\frac{1}{z})$ and $sin\theta = -\frac{1}{2i}(z-\frac{1}{z})$

my residue simplifies to $-\frac{(z-z_0)(z^2-1)}{z(6z+2z^2+2)}$

At $z_0 = 0$

I get

$Res[f,0] = 1/2$

but my other two roots, of the polynomial in the denominator, are giving me problems when trying to compute the residue.

$\frac{1}{2}(\sqrt5-3)$ and $\frac{1}{2}(-3-\sqrt5)$

I think both these points are in my region? it's a full revolution yes?

 

[KleZMeR]

I'm trying LHopital

 

[KleZMeR]

Seemed to have worked! With my root that is within the circle $\frac{1}{2}(\sqrt5-3)$ I get $\frac{1}{2}$

And so $\frac{1}{2} + (-\frac{1}{2}) = 0 = Res[f,0] + Res[f,\frac{1}{2}(\sqrt5-3)]$

 

[vela]

my residue simplifies to $-\frac{(z-z_0)(z^2-1)}{z(6z+2z^2+2)}$

I wouldn't call that the residue. You mean the residue for a simple pole at $z=z_0$ is given by
$$b_1 = \lim_{z \to z_0} -\frac{(z-z_0)(z^2-1)}{z(6z+2z^2+2)}.$$

but my other two roots, of the polynomial in the denominator, are giving me problems when trying to compute the residue.

I still don't know what problem you ran into.