Alright this one should be easy, but I do not really see how can I take advantage of the principle of the argument here, no matter how I arrange the terms.
$$\int_C \frac{e^z}{\cosh z}\,dz$$
The contour $C$ is the circumference of center $0$ and radius $5$, but I am more interested in how the manipulation to use the Argument Principle goes than the actual result.
A hint please?
For the argument principle, you want the integrand to be of the form
$$\frac{f'(z)}{f(z)}$$
(modulo constant factors).
Here we have $\cosh z$ in the denominator, and its derivative is $\sinh z = \frac{1}{2}(e^z - e^{-z})$, thus you want to introduce some $e^{-z}$. Since $C$ is a circle with centre $0$, if $z$ traverses $C$, then so does $-z$, in the same direction, just with different start and end point. So we have
$$\int_C \frac{e^z}{\cosh z}\,dz = \int_C \frac{e^{-z}}{\cosh (-z)}\,d(-z) = -\int_C \frac{e^{-z}}{\cosh z}\,dz$$
since $\cosh$ is an even function. Adding both forms of the integral and dividing by $2$, we obtain
$$\int_C \frac{e^z}{\cosh z}\,dz = \frac{1}{2}\int_C \frac{e^z - e^{-z}}{\cosh z}\,dz = \int_C \frac{\sinh z}{\cosh z}\,dz.$$
The rest is a matter of counting zeros.