Evaluate $\int_\gamma \frac{e^z}{(z^2+1)^2} dz$ where $\gamma$ is $C(0,2)$ traversed twice counter-clockwise. This $\gamma$ should be represented as a cycle I believe.
Hi all, I am wondering about how to evaluate this integral using the Cauchy Integral Formula.
I have tried to represent the integral as a linear combination of simpler integrals but I'm unsure how. Thank you.
Hint. Given partial fraction expansion $$\frac{1}{(z^2+1)^2}=-\frac{1}{4(z-i)^2}-\frac{i}{4(z-i)}-\frac{1}{4(i+z)^2} + \frac{i}{4(i + z)}$$ Cauchy's integral formula can be applied for each part individually, considering the winding number of course.
$$\int\limits_{\gamma}\frac{e^z}{(z^2+1)^2} dz=-\int\limits_{\gamma}\frac{e^z}{4(z-i)^2}dz-\int\limits_{\gamma}\frac{ie^z}{4(z-i)}dz-\int\limits_{\gamma}\frac{e^z}{4(i+z)^2}dz + \int\limits_{\gamma}\frac{ie^z}{4(i + z)}dz$$
More details here (section 6) and here (pages 9 and 12).