Determine \(\int_\gamma \frac{dz}{z^3\cos z}\), where \(\gamma\) is the circle \(\abs{z-1}=2\) traversed counterclockwise.
Erratum: the source writes the contour as “the circle \(\{\abs{z-1}<2\}\)”. That set is the open disk; the contour is its boundary circle \(\abs{z-1}=2\).
Solution
1. The only enclosed poles are \(0\) and \(\pi/2\), with no pole on the contour.
Put \(F(z)=1/(z^3\cos z)\). Besides zero, its possible poles are the zeros of cosine. The equation \(\cos z=0\) is equivalent to \(e^{2iz}=-1\), hence to \(z=\pi/2+k\pi\) for \(k\in\ZZ\): taking moduli first forces \(\operatorname{Im}z=0\), and the real solutions are the indicated odd multiples of \(\pi/2\).
A real point is inside the contour exactly when it belongs to \((-1,3)\). The bounds \(2<\pi<4\) follow from \(\pi=4\int_0^1(1+t^2)^{-1}\,dt\) and \(1/2<(1+t^2)^{-1}<1\) for \(0<t<1\). Thus \(\pi/2\in(1,2)\) lies inside, while \(-\pi/2<-1\) and \(3\pi/2>3\); all other cosine zeros are farther away. Also \(\abs{0-1}=1<2\). These strict inequalities exclude boundary poles and identify every enclosed pole.
2. The integral is\[ \boxed{\int_\gamma \frac{dz}{z^3\cos z}=i\pi-\frac{16i}{\pi^2}}. \]
At zero the Taylor expansion gives \[ \cos z=1-\frac{z^2}{2}+O(z^4),\qquad F(z)=\frac1{z^3}+\frac1{2z}+O(z). \] Therefore \(\operatorname{Res}_0F=1/2\). At \(a=\pi/2\), cosine has a simple zero with derivative \(-\sin a=-1\), so \[ \operatorname{Res}_{a}F =\frac1{a^3(-\sin a)}=-\frac8{\pi^3}. \] The residue theorem applies to the meromorphic function \(F\) on the disk and its counterclockwise boundary. Using step 1, it gives \[ \int_\gamma F(z)\,dz =2\pi i\left(\frac12-\frac8{\pi^3}\right) =i\pi-\frac{16i}{\pi^2}. \]
3. Q.E.D.
Step 2 gives the requested integral.