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Let a be a real number with |a|>1. Compute $$\int_0^{2\pi}\frac{ 1-a\cos\theta}{1-2a\cos\theta + a²} dθ$$

I know i should think of a circle since the bounds are from 0 to 2π. I have the soltuions to this question. But what i don't understand is how do we assume that we should start the problem with

$$∫_{|z|=1} \frac{dz}{z-a}dz\,\;\; ?????\;\;\;\text{How do i assume this at the start???} $$

and then use $\,\,z=e^{iθ}\,\,$ and $\,\,dz= ie^{iθ}\,\,$ to get a similar integral as the one above after doing some algebra.

I have not learnt the Residual formula yet. I have only learnt until the Cauchy Integral Formula

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Is acos = a*cos or arccos? Please consider giving $\LaTeX$ format to your question. –  Pragabhava Oct 24 '12 at 18:12
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Check whether my editing is accurate...and please learn how to use LaTeX for mathematics in this site! –  DonAntonio Oct 24 '12 at 18:16
    
Related: math.stackexchange.com/questions/211058/… –  Pragabhava Oct 24 '12 at 18:18

3 Answers 3

If $z=e^{i\theta}$ then $$\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}=\frac{z+\frac1z}{2}=\frac{z^2+1}{2z}$$ Hence $$\begin{align*} \int_0^{2\pi}\frac{ 1-a\cos\theta}{1-2a\cos\theta + a²} d\theta=&\int_{|z|=1}\frac{1-a\frac{z^2+1}{2z}}{1-2a\frac{z^2+1}{2z}+a^2}\frac{dz}{iz}=\int_{|z|=1}\frac{2z-az^2-a}{2iz(z-az^2-a+a^2z)}dz\\ =& \int_{|z|=1}\frac{-az^2+2z-a}{-2iaz(z^2-(a+\frac1a)z+1)}dz=\frac{1}{2ia}\int_{|z|=1}\frac{az^2-2z+a}{z(z-a)(z-\frac1a)}dz \\ \end{align*}$$ You know that $|a|>1$, $\left|\frac1a\right|<1$. Can you continue from here?

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It's not at all obvious. But a clue is that by the Law of Cosines $1 - 2 a \cos \theta + a^2$ is the square of the third side of a triangle with two sides $1$ and $a$ and angle $\theta$ between them. Thus it is $|e^{i\theta} - a|^2$. If $z=e^{i\theta}$ is on the unit circle, $|z - a|^2 = (z - a)(\overline{z} - a) = (z - a)(1/z - a)$. On the other hand, $1 - a \cos(\theta) = 1 - \text{Re}(a z) = 1 - (a z + a/z)/2$, and $d\theta = dz/(iz)$.

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Let assume that $\gamma:[0,2\pi]\to\mathbb{R}^2$ is a closed and differentiable curve. Since $$\frac{d}{d\theta}\arctan\left(\frac{y(\theta)}{x(\theta)}\right)=\frac{x\,y'-y\,x'}{x^2+y^2}$$ we have that $$\int_{0}^{2\pi}\frac{x\,y'-y\,x'}{x^2+y^2}\,dt$$ is the topological degree of $\gamma$ wrt the origin, i.e. $2\pi$ times the winding number of $\gamma$. If we take: $$ \gamma(\theta) = \left(1-\frac{1}{a}\cos(\theta),\frac{1}{a}\sin(\theta)\right),$$ that is a circle not enclosing the origin, we immediately have that the value of the integral in the exercise is zero.

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