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Homework Statement
Evaluate [itex]\oint[/itex][itex]\frac{1}{a-cos\theta}[/itex] d[itex]\theta[/itex] from [0:2[itex]\pi[/itex]]
Homework Equations
Method of contour integration.
The Attempt at a Solution
I make the substitution z=exp(i[itex]\theta[/itex]) [itex]\Rightarrow[/itex]
giving me the integral:
[itex]\oint[/itex]1/(a-½(z+z-1)) dz/iz from [0:2[itex]\pi[/itex]]
And rearranging:
2/i [itex]\oint[/itex]1/(-z2+2az-1) dz from [0:2[itex]\pi[/itex]]
The quadratatic equation in the denominator has the roots z0 = a ± √(a2-1). Only the one with negative sign is inside the unit circle for a>1 (actually, how can you know that it will be that for all a>1?) which gives:
Pole at z0 = a - √(a2-1)
So we can write f(z) as:
f(z) = 2i/((z-a + √(a2-1))(z-a - √(a2-1)))
And since it's a simple pole we're dealing with the residue can be found as:
limz->z0{(z-z0)f(z)} = limz->a - √(a2-1){(z-a+√(a2-1) 2i/((z-a + √(a2-1)(z-a - √(a2-1))} = limz->a - √(a2-1){2i/(z-a - √(a2-1)} = 2i/-2√(a2-1) = -i/√(a2-1)
And the residue theorem gives the final value for the integral, I:
I = 2[itex]\pi[/itex]/√(a2-1)
Is this correct?
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