What is the value of the corresponding contour integral?

[bugatti79]

Homework Statement

Convert the following to an equivalent cotour integral around |z|=1 then use Cauchy's integral formula to evaluate it.

$\int_{0}^{2 \pi} \frac {d \theta}{13+5 \sin \theta}$

Homework Equations

let $z=e^{i \theta}$

The Attempt at a Solution

$d \theta = \frac{dz}{i z}$

$\displaystyle \int_{0}^{2 \pi} \frac {d \theta}{13+5 \sin \theta}=\int_{|z|=1} \frac {dz/iz}{13+5/2i(e^{i \theta} -e^{-i \theta})}\frac{2iz}{2iz}$

$\displaystyle =\int_{|z|=1} \frac {2dz}{26iz+5z^2-5}$

where the denominator has the roots $i(2.6 \pm 2.4)$ using quadratic formula...so far ok?

 

[jackmell]

Homework Statement

Convert the following to an equivalent cotour integral around |z|=1 then use Cauchy's integral formula to evaluate it.

$\int_{0}^{2 \pi} \frac {d \theta}{13+5 \sin \theta}$

Homework Equations

let $z=e^{i \theta}$

The Attempt at a Solution

$d \theta = \frac{dz}{i z}$

$\displaystyle \int_{0}^{2 \pi} \frac {d \theta}{13+5 \sin \theta}=\int_{|z|=1} \frac {dz/iz}{13+5/2i(e^{i \theta} -e^{-i \theta})}\frac{2iz}{2iz}$

$\displaystyle =\int_{|z|=1} \frac {2dz}{26iz+5z^2-5}$

where the denominator has the roots $i(2.6 \pm 2.4)$ using quadratic formula...so far ok?

The integral is ok but I get different signs on the roots. Maybe double check them.

 

[bugatti79]

Homework Statement

Convert the following to an equivalent cotour integral around |z|=1 then use Cauchy's integral formula to evaluate it.

$\int_{0}^{2 \pi} \frac {d \theta}{13+5 \sin \theta}$

Homework Equations

let $z=e^{i \theta}$

The Attempt at a Solution

$d \theta = \frac{dz}{i z}$

$\displaystyle \int_{0}^{2 \pi} \frac {d \theta}{13+5 \sin \theta}=\int_{|z|=1} \frac {dz/iz}{13+5/2i(e^{i \theta} -e^{-i \theta})}\frac{2iz}{2iz}$

$\displaystyle =\int_{|z|=1} \frac {2dz}{26iz+5z^2-5}$

where the denominator has the roots $i(2.6 \pm 2.4)$ using quadratic formula...so far ok?

The integral is ok but I get different signs on the roots. Maybe double check them.

Yes, you are right, should be $(-2.6 \pm 2.4)i$

continuing on

$|(-2.6 +2.4)i| < 1 \implies (-2.6 +2.4)i$ lies inside $|z|=1$
$|(-2.6 -2.4)i| > 1 \implies (-2.6 -2.4)i$ lies outside $|z|=1 \therefore$

$\displaystyle \int_{|z|=1} \frac {2dz}{26iz+5z^2-5}=2 \pi i f(-2.6+2.4)i=$

$\displaystyle 2 \pi i \frac{2}{(-2.6-2.4)i-(-2.6+2.4)i}=-\frac{\pi}{1.2}$...?