How I Impressed My Wife: Part 4g

So far in this series, I have used three different techniques to show that

Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.

For the third technique, a key step in the calculation was showing that the residue of the function

f(z) = \displaystyle \frac{1}{z^2 + 2\frac{S}{R}z + 1} = \displaystyle \frac{1}{(z-r_1)(z-r_2)}

at the point

r_1 = \displaystyle \frac{-S + \sqrt{S^2 -R^2}}{R}

was equal to

\displaystyle \frac{R}{ 2 \sqrt{S^2-R^2} }.

Initially, I did this by explicitly computing the Laurent series expansion about z = r_1 and identifying the coefficient for the term (z-r_1)^{-1}.

In this post and the next post, I’d like to discuss alternate ways that this residue could have been obtained.
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Notice that the function f(z) has the form \displaystyle \frac{g(z)}{(z-r) h(z)}, where g and h are differentiable functions so that g(r) \ne 0 and h(r) \ne 0. Therefore, we may rewrite this function using the Taylor series expansion of \displaystyle \frac{g(z)}{h(z)} about z = r:

f(z) = \displaystyle \frac{1}{z-r} \left[ \frac{g(z)}{h(z)} \right]

f(z) = \displaystyle \frac{1}{z-r} \left[ a_0 + a_1 (z-r) + a_2 (z-r)^2 + a_3 (z-r)^3 + \dots \right]

f(z) = \displaystyle \frac{a_0}{z-r} + a_1 + a_2 (z-r) + a_3 (z-r)^2 + \dots

Therefore, the residue at z = r is equal to a_0, or the constant term in the Taylor expansion of \displaystyle \frac{g(z)}{h(z)} about z = r. Therefore,

a_0 = \displaystyle \frac{g(r)}{h(r)}

For the function at hand g(z) \equiv 1 and h(z) = z-r_2. Therefore, the residue at z = r_1 is equal to \displaystyle \frac{1}{r_1 - r_2}, matching the result found earlier.

 

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  1. How I Impressed My Wife: Index | Mean Green Math

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