Complex hyperbolic Trigonometry

Question

When faced with the equation

$\cos{z}=\sqrt{2}$

I want to solve for z so I break it up into a sum $z=x+iy$ and get:

$\cos{z}=\cos{x}\cosh{y}-i \sin{x} \sinh{y}$

equating real and imaginary parts I am faced with

$\cos{x}\cosh{y}=\sqrt{2}$ and $\sin{x}\sinh{y}=0$

How do I go about solving from here I can't seem to get out of this loop where I have to end up using some ugly form for inverse hyperbolic function.

EDIT: $\cos^{-1}{z}$ is defined in Churchill as

$\cos^{-1}{z}=-i\log{[z+i(1-z^2)^{1/2}]}$

Am I better off just plugging it in to here?

Answer 1

Avoid squaring whenever possible as it immediately introduces extraneous root(s)

We have $\displaystyle\frac{e^{iz}+e^{-iz}}2=\sqrt2$

$$\iff(e^{iz})^2-2\sqrt2(e^{iz})+1=0$$

$$\implies e^{iz}=\dfrac{2\sqrt2\pm\sqrt{8-4}}2=\sqrt2\pm1$$

$$\iff iz=\log(\sqrt2\pm1)$$

$$\iff z=-i\log(\sqrt2\pm1)$$

Answer 2

First separate to cases $y=0$ and $y \neq 0$. You will find that the first has no solution. The latter implies $x=k\pi$ that implies $y = (-1)^k \textrm{arcosh}(\sqrt{2}) = (-1)^k \ln(\sqrt{2} + 1)$.

Answer 3

Squaring both sides you'll get $$\cos^2(z) = 2 \\ \implies 1-\sin^2(z) = 2 \\ \implies \sin^2(z) = -1 \\ \implies \sin(z) = \pm i \\ \implies z = \arcsin(\pm i)$$ You can see this works by the identity $$\cos(\arcsin(x)) = \sqrt{1-x^2}$$ (assuming the identity holds for complex numbers) And you could convert $\arcsin(\pm i)$ into $$\arcsin(\pm i) = -i\ln\left(i(\pm i)+\sqrt{1-(\pm i)^2} \right)$$