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 $\displaystyle\frac{e^{iz}+e^{-iz}}2=\sqrt2$
$$\iff(e^{iz})^2-2\sqrt2(e^{iz})+1=0$$
$$\implies e^{iz}=\dfrac{2\sqrt2\pm\sqrt{8-4}}2=\sqrt2\pm1$$
$$\iff iz=\log(\sqrt2\pm1)$$
$$\iff z=-i\log(\sqrt2\pm1)$$
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