Calculate the integral $\cos(x) \cosh(y) + i \sin(x) \sinh(y)$

Question

Given $g(z) = \cos(x) \cosh(y) + i \sin(x) \sinh(y)$ and $$W = \int_{{\textstyle\frac{\pi}{2}} + i \log(2)}^{{\textstyle\frac{\pi}{2}} + i \log (5)} \frac{\mathrm{d}z}{g(z)} = \int_{{\textstyle\frac{\pi}{2}} + i \log(2)}^{{\textstyle\frac{\pi}{2}} + i \log (5)} \frac{\mathrm{d}z}{\cos(x) \cosh(y) + i \sin(x) \sinh(y)},$$ how to calculate $W$? I do not know how to start…

Answer 1

Taking $z = x+iy$ and because $\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$ then $$\cos(z) = \cos(x)\cos(iy) - \sin(x)\sin(iy)$$ also $\sin(iy) = i\sinh(y)$ and $\cos(iy) = \cosh(y)$ $$\cos(z) = \cos(x)\cosh(y) - i\sin(x)\sinh(y)$$ let's assume $\tilde{x} = -x$, if $\cos(x) = \cos(\tilde{x})$ but $\sin(x) = -\sin(\tilde{x})$ so being $z = -\tilde{x}+iy$ $$\cos(z) = \cos(\tilde{x})\cosh(y) + i\sin(\tilde{x})\sinh(y)$$ from there the integral is known $$W = \int\frac{1}{g(z)}~dz = \int\frac{1}{\cos(-x+iy)}~dz = \int\frac{1}{\cos(z)}~dz = \ln\left(\frac{1}{\cos(z)}+\tan(z)\right)$$