Evaluate \[ I=\frac1{2\pi i}\int_{|z|=1}\frac{(z+2)^2}{z^2(2z-1)}\,dz, \] where the circle is oriented counterclockwise.
Solution
Set \[ F(z)=\frac{(z+2)^2}{z^2(2z-1)}. \]
1. The poles of \(F\) inside \(\abs{z}=1\) are a double pole at \(0\) and asimple pole at \(1/2\).
The denominator is \(z^2(2z-1)\). Its zeros are \(0\), with multiplicity \(2\), and \(1/2\), with multiplicity \(1\). The numerator is nonzero at both points, and both have modulus less than \(1\).
2.\[ \operatorname{Res}(F,0)=-12. \]
For the double pole at \(0\), \[ \operatorname{Res}(F,0) = \left. \frac{d}{dz} \left(\frac{(z+2)^2}{2z-1}\right) \right|_{z=0}. \] Differentiating gives \[ \frac{d}{dz} \left(\frac{(z+2)^2}{2z-1}\right) = \frac{2(z+2)(2z-1)-2(z+2)^2}{(2z-1)^2} = \frac{2(z+2)(z-3)}{(2z-1)^2}. \] At \(z=0\) this equals \(-12\).
3.\[ \operatorname{Res}(F,1/2)=25. \]
Since \(2z-1=2(z-1/2)\), \[ \operatorname{Res}(F,1/2) = \left. \frac{(z+2)^2}{2z^2} \right|_{z=1/2} = \frac{(5/2)^2}{2(1/2)^2} =25. \]
4.\[ I=\boxed{13}. \]
By the residue theorem and steps <1>1–<1>3, \[ I = \frac{1}{2\pi i}\int_{\abs{z}=1}F(z)\,dz = \operatorname{Res}(F,0)+\operatorname{Res}(F,1/2) = -12+25 =13. \]
5. Q.E.D.
Step 4 gives the requested value.