Evaluate \[ \int_C\frac{e^z}{z(2z+1)^2}\,dz, \] where \(C\) is the unit circle oriented counterclockwise.
Solution
Set \[ R(z)\coloneqq\frac{e^z}{z(2z+1)^2}. \]
1. The poles of \(R\) inside \(C\) are a simple pole at \(z=0\) and a double pole at \(z=-1/2\).
The denominator vanishes only at \[ z=0 \qquad\text{and}\qquad z=-\frac12. \] Both points lie in \(\abs z<1\). The factor \(z\) occurs to the first power, while \[ (2z+1)^2=4(z+1/2)^2. \]
2. The residue at \(z=0\) is\[ \operatorname{Res}(R;0)=1. \]
Since the pole is simple, \[ \operatorname{Res}(R;0) = \lim_{z\to0}\frac{e^z}{(2z+1)^2} =1. \]
3. The residue at \(z=-1/2\) is\[ \operatorname{Res}(R;-1/2) = -\frac32e^{-1/2}. \]
Using \[ R(z) = \frac{1}{(z+1/2)^2}\frac{e^z}{4z}, \] the residue at the double pole is \[ \left.\frac{d}{dz}\left(\frac{e^z}{4z}\right)\right|_{z=-1/2}. \] Now \[ \frac{d}{dz}\left(\frac{e^z}{4z}\right) = \frac14e^z\left(\frac1z-\frac1{z^2}\right), \] so at \(z=-1/2\) this equals \[ \frac14e^{-1/2}(-2-4) = -\frac32e^{-1/2}. \]
4. The integral is\[ \boxed{ 2\pi i\left(1-\frac32e^{-1/2}\right)}. \]
By the residue theorem and steps <1>2–<1>3, \[ \begin{aligned} \int_C R(z)\,dz &=2\pi i\left( \operatorname{Res}(R;0) +\operatorname{Res}(R;-1/2) \right)\\ &=2\pi i\left(1-\frac32e^{-1/2}\right). \end{aligned} \]
5. Q.E.D.
Step 4 gives the requested value.