Use residues at infinity to evaluate \[ \int_{|z|=1} \frac{1}{(z-2)(1+2 z)^{4}(1-3 z)^{7}} \dz .\]
Solution
Goal. Evaluate \(\oint_{|z|=1} \frac{1}{(z-2)(1+2z)^4(1-3z)^7}\,dz\) using residues at infinity.
1. The integrand \(f(z) = \frac{1}{(z-2)(1+2z)^4(1-3z)^7}\) has poles inside \(|z| = 1\) at \(z = -1/2\) (order \(4\)) and \(z = 1/3\) (order \(7\)), and a simple pole at \(z = 2\) outside.
\(1 + 2z = 0\) at \(z = -1/2\) and \(1 - 3z = 0\) at \(z = 1/3\), both inside the unit circle; \(z = 2\) is outside.
2. The sum of all residues over the extended plane is \(0\).
for a rational function, \(\sum_{\text{finite poles}} \operatorname{Res}(f, z_j) + \operatorname{Res}(f, \infty) = 0\).
3. Compute \(\operatorname{Res}(f, \infty)\).
3.1. \(\operatorname{Res}(f, \infty) = -\operatorname{Res}\qty(\frac{1}{z^2} f(1/z), 0)\).
definition of the residue at infinity.
3.2. \(\frac{1}{z^2} f(1/z) = \frac{1}{z^2} \cdot \frac{1}{(1/z - 2)(1 + 2/z)^4(1 - 3/z)^7} = \frac{z^{10}}{(1-2z)(z+2)^4(z-3)^7}\).
substitute \(z \mapsto 1/z\) and simplify (multiply numerator and denominator by \(z^{12}\)).
3.3. This is analytic at \(z = 0\) (numerator has \(z^{10}\)), so \(\operatorname{Res}\qty(\frac{1}{z^2}f(1/z), 0) = 0\).
no \(1/z\) term in the Laurent expansion at \(0\).
3.4. Hence \(\operatorname{Res}(f, \infty) = 0\).
3.1 and 3.3.
4. Compute \(\operatorname{Res}(f, 2)\).
4.1. \(z = 2\) is a simple pole, so \(\operatorname{Res}(f, 2) = \lim_{z \to 2} (z-2) f(z) = \frac{1}{(1+2\cdot 2)^4 (1 - 3\cdot 2)^7} = \frac{1}{5^4 \cdot (-5)^7} = -\frac{1}{5^{11}}\).
the residue at a simple pole is the limit of \((z-2)f(z)\).
5. Sum of residues inside \(|z| = 1\).
5.1. \(\sum_{\text{inside}} \operatorname{Res} = -(\operatorname{Res}(f, 2) + \operatorname{Res}(f, \infty))\).
by 2, the total is zero, and the poles outside are \(z = 2\) and \(\infty\).
5.2. \(= -\qty(-\frac{1}{5^{11}} + 0) = \frac{1}{5^{11}}\).
substitute 3.4 and 4.1.
6. \(\oint_{|z|=1} f(z)\,dz = 2\pi i \cdot \frac{1}{5^{11}}\).
the residue theorem.
7. Q.E.D.
the integral equals \(\frac{2\pi i}{5^{11}}\).