Calcular la integral $\cos(x) \cosh(y) + i \sin(x) \sinh(y)$

Question

Dado $g(z) = \cos(x) \cosh(y) + i \sin(x) \sinh(y)$ y $$W = \int_{{\textstyle\frac{\pi}{2}} + i \log(2)}^{{\textstyle\frac{\pi}{2}} + i \log (5)} \frac{\mathrm{d}z}{g(z)} = \int_{{\textstyle\frac{\pi}{2}} + i \log(2)}^{{\textstyle\frac{\pi}{2}} + i \log (5)} \frac{\mathrm{d}z}{\cos(x) \cosh(y) + i \sin(x) \sinh(y)},$$ cómo calcular $W$ ? No sé cómo empezar

Answer 1

Tomando $z = x+iy$ y porque $\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$ entonces $$\cos(z) = \cos(x)\cos(iy) - \sin(x)\sin(iy)$$ también $\sin(iy) = i\sinh(y)$ y $\cos(iy) = \cosh(y)$ $$\cos(z) = \cos(x)\cosh(y) - i\sin(x)\sinh(y)$$ supongamos que $\tilde{x} = -x$ , si $\cos(x) = \cos(\tilde{x})$ pero $\sin(x) = -\sin(\tilde{x})$ por lo que ser $z = -\tilde{x}+iy$ $$\cos(z) = \cos(\tilde{x})\cosh(y) + i\sin(\tilde{x})\sinh(y)$$ a partir de ahí se conoce la integral $$W = \int\frac{1}{g(z)}~dz = \int\frac{1}{\cos(-x+iy)}~dz = \int\frac{1}{\cos(z)}~dz = \ln\left(\frac{1}{\cos(z)}+\tan(z)\right)$$