Complex hyperbolic Trigonometry

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

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

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

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

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

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

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