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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?

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    Is it $\displaystyle\int_\gamma \frac{e^z}{\cosh z}\,dz$, where $\gamma$ is the said circle? – 2017-01-16
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    @egreg Yes, I'll edit to make it clear. – 2017-01-16
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    The zeros of $\cosh z$ are at $(\frac{\pi}{2}+k\pi)i$; find those in the given circle and compute the residues at them of $e^z/\cosh z$, which has poles where $\cosh z$ has zeros. – 2017-01-16

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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.