Show that the following integral is equal to zero by evaluating $\oint_c e^{z} dz = 0$ $$\int_{0}^{2\pi} e^{\cos \theta } \cos(\theta + \sin\theta) d\theta = \int_{0}^{2\pi} e^{\cos \theta } \sin(\theta + \sin\theta) d\theta = 0$$
I think this is equivalent to showing $\oint_c e^{z + i \arg (z)} dz = 0$ but I don't know how :((
Let $z=e^{i \theta}$, $dz = i e^{i \theta} d\theta$.
$$\begin{align}\oint_C dz \: e^z &= i \int_0^{2 \pi} d\theta \: e^{i \theta} e^{e^{i \theta}}\\ &= i \int_0^{2 \pi} d\theta \: e^{i \theta} e^{\cos{\theta} + i \sin{\theta}}\\ &= i \int_0^{2 \pi} d\theta \: e^{\cos{\theta}} (\cos{(\theta+\sin{\theta})} + i \sin{(\theta+\sin{\theta})})\end{align}$$
Because $e^z$ is analytic within the region inside the contour $C$, the integral is zero. Therefore, the real and imaginary parts of that last integral are zero.