Evaluate: $$\int_{\gamma} \frac{e^{\cos z}}{(z^2 + 1)^2} dz$$ Where $\gamma = \omega_1 * \omega_2$ and $\omega_1 = C(-1 + i, \sqrt{2})$ with positive orientation, $\omega_2 = C(1 - i, \sqrt{2})$ with negative orientation. Both circles have their beginning and end in $0$.
That looks easy since there may be only $1$ point of singularity of the integrand inside of $\gamma$ since the only such place can be in $z = i$. However the definition of $\gamma$ is somewhat complicated because it is made of $2$ circles with different orientations. I don't know how to approach such a problem.
$\newcommand{\d}{\,\mathrm{d}}\newcommand{\res}{\operatorname{Res}}$I am going to assume by $\omega_1\star\omega_2$ you mean the concatenated contour that follows $\omega_2$ in the first half of the time interval, beginning and ending at the origin, and follows $\omega_1$ in the second half - also beginning and ending at the origin.
Now the residue theorem really says (think about the proof - it essentially reduces integration of $f$ to integration about the $(-1)$th components of the Laurent series, and these integrals are just residue $\times$ winding number) : $$\oint_\gamma f(z)\d z=2\pi i\cdot\sum_{\zeta\,\text{ is a pole }}\res_{z=\zeta}(f)\cdot N_\gamma(\zeta)$$
Where the winding number appears. This is often omitted because contours with winding number $1$ are typically used. Our $f$ is holomorphic, bar the two poles at $\zeta=\pm i$. Our $\gamma$ is a closed contour, so the residue theorem applies.
What is $N_\gamma(i)$? There is a direct formula for the winding number, namely: $$\frac{1}{2\pi i}\oint_\gamma\frac{1}{z-i}\d z=\frac{1}{2\pi i}\oint_{\omega_2}\frac{1}{z-i}\d z+\frac{1}{2\pi i}\oint_{\omega_1}\frac{1}{z-i}\d z$$See, the winding number behaves linearly w.r.t path concatenation. The winding is $1$ in one component, and nil in the other. When dealing with $\omega_2$, as it has reversed orientation the winding about $-i$ is $-1$. $N_\gamma(\pm i)$ is therefore $\pm1$ so we get: $$\oint_\gamma f(z)\d z=2\pi i\cdot\left[\res_{z=i}(f)-\res_{z=-i}(f)\right]$$
Which you can compute from here.