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if C is the positively oriented circle $ \lvert z \rvert = 4 $

$cot^2z = \frac{cos^2z}{sin^2z}$

singularites will occur when $sin^2z = 0$, so $\pi$ and $ -\pi $ within $\lvert z \rvert = 4$

I have tried using my p over q rule

$p(z_0) = \cos^2z \neq 0 $ at either singularity

$q(z_0) = \sin^2z = 0 $ at both singularities, but

$ q^{'}(z_0) = 2\cos(z)\sin(z) = 0 $ at all my singularities

I can't use my $ \phi $-rule because $\cos^2z$ is analytic and but I can't get $ \sin^2z $ into the for of $(z-z_0)^{-m} $

Am I looking in the wrong place trying to use Residues at the two singularies to calculate the integral?

guyman
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1 Answers1

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The residue at $z=z_0$ is the coefficient of the $(z-z_0)^{-1}$ term in the Laurent expansion of the function about $z=z_0$. We know that the poles of $\cot^2z$ are of order $2$, so the Laurent expansion is of the form $\sum_{k=-2}^\infty A_k(z-z_0)^k$. Now $\sin z = z + O(z^3)$ so $\sin^2z= z^2 + ^O(z^4)$ and we see that $\sin^2z\cot^2z$ has a linear term of $a_1z$. But $\sin^2z\cot^2z=\cos^2z$, which has no linear term; that is, the residue of $\sin z$ at $z=0$ is $0$. The same argument applies at $\pi$ and $-\pi$, so the integral is $0$.

saulspatz
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