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Contour Integration - my solution for real integral is complex?

So I've had a crack at this contour integration question and have somehow managed to get a complex solution for a real integral... I've gone through my working a number of times but can't seem to find the mistake, so was wondering if anyone on TSR could help.

The Question

Evaluate the integral

[br]I=\int^\pi_{-\pi} \frac{\,d\theta}{a+b\cos\theta+c\sin\theta}[br]

where

a

b

c

are real positive constants such that

a^2>b^2+c^2>0

My Attempt

Consider the complex function

f(z)=\frac{1}{az+\frac{1}{2}b(z^2+1)+\frac{1}{2i}c(z^2-1)}

This has simple poles when denominator equals zero, i.e. at

z=z_1

and

z=z_2

, where

z_1=\frac{-a+\sqrt{a^2-(b^2+c^2)}}{b-ic}

and

z_2=\frac{-a-\sqrt{a^2-(b^2+c^2)}}{b-ic}

by the quadratic formula. See that

|z_2|>|z_1|\,\,\,\,\,\,(*)

.

Note that

|z_1||z_2|=|z_1 z_2|=\left|\frac{b^2+c^2}{(b-ic)^2}\right|=1

Referring back to

(*)

we see that

|z_1|<1

and

|z_2|>1

\text{Res}[f(z), z_1]=lim_{z\to z_1}[(z-z_1)f(z)]=\frac{1}{z_1-z_2}=\frac{b-ic}{2 \sqrt{a^2-b^2-c^2}}

Since

z_1

is the only pole enclosed by the contour

|z|=1

, by the residue theorem:

\oint_{|z|=1}f(z)dz=2\pi i \text{Res}[f(z), z_1]\,\,\,\,\,\,(**)

Along the contour

|z|=1

, we can write

z=e^{i\theta}

dz=ie^{i\theta}d\theta

with

\theta

[-\pi,\pi]

Then

\oint_{|z|=1}f(z)dz=\int^\pi_{-\pi}\frac{ie^{i\theta}}{ae^{i \theta}+\frac{1}{2}b(e^{2 i \theta}+1)+\frac{1}{2i}c(e^{2 i \theta}-1)}d\theta=i\int^\pi_{-\pi}\frac{1}{a+\frac{1}{2}b(e^{i\theta}+e^{-i \theta})+\frac{1}{2i}c(e^{i \theta}-e^{-i \theta})}d\theta=iI

Returning to

(**)

, we see that

I=2\pi\text{Res}[f(z), z_1]=\frac{b-ic}{\sqrt{a^2-b^2-c^2}}\pi

Why is my result complex? I'd appreciate it if someone could point out where I went wrong.

(edited 10 years ago)

Reply 1

ghostwalker

Original post by IchiCC

Note that

=\left|\frac{b^2+c^2}{(b-ic)^2}\right|=1

Can't say I know anything about contour integration ( :sad:

), however I can't see the justification for that equalling 1.

Refering back to the original quadratic, the product of the two roots is:

~~b/2 + ci/2~~ Rubbsh.

Edit: And why am I awake at 2 o'clock in the morning?