Evaluate \[ \frac{1}{2\pi i}\int_{|z|=1}\frac{z^{11}}{12z^{12}-4z^9+2z^6-4z^3+1}\,dz, \] where the direction of integration is counterclockwise.
Solution
Let \[ P(z)=12z^{12}-4z^9+2z^6-4z^3+1 \] and \[ R(z)=\frac{z^{11}}{P(z)}. \]
1. Every zero of \(P\) lies in the open unit disk.
On \(\abs z=1\), \[ \abs{12z^{12}}=12, \] while \[ \begin{aligned} \abs{-4z^9+2z^6-4z^3+1} &\leq 4+2+4+1\\ &=11. \end{aligned} \] Thus \[ \abs{-4z^9+2z^6-4z^3+1} < \abs{12z^{12}} \] on the unit circle. By Rouché’s theorem, \(P\) and \(12z^{12}\) have the same number of zeros in the unit disk, counted with multiplicity. Since \(P\) has degree \(12\), all twelve of its zeros lie there. In particular, \(P\) has no zero on the unit circle.
2. The residue of \(R\) at infinity is\[ \operatorname{Res}_{\infty}R=-\frac1{12}. \]
As \(z\to\infty\), \[ \begin{aligned} R(z) &= \frac{z^{11}}{12z^{12}-4z^9+2z^6-4z^3+1}\\ &= \frac1{12z}+O\left(\frac1{z^4}\right). \end{aligned} \] Therefore the coefficient of \(z^{-1}\) in the Laurent expansion of \(R\) at infinity is \(1/12\). By the definition of the residue at infinity, \[ \operatorname{Res}_{\infty}R=-\frac1{12}. \]
3. The sum of the residues of \(R\) at all of its finite poles is\[ \frac1{12}. \]
For a rational function, the sum of all finite residues together with the residue at infinity is zero. Hence step 2 gives \[ \sum_{a\in\CC}\operatorname{Res}_{z=a}R = -\operatorname{Res}_{\infty}R = \frac1{12}. \]
4. The required integral is\[ \boxed{\frac1{12}}. \]
By step 1, every finite pole of \(R\) lies inside \(\abs z=1\), and none lies on the contour. The residue theorem and step 3 therefore give \[ \frac1{2\pi i}\int_{\abs z=1}R(z)\,dz = \sum_{\abs a<1}\operatorname{Res}_{z=a}R = \frac1{12}. \]
5. Q.E.D.
Step 4 is the required evaluation.