How do I show $\arccos(4)$ in form $a+bi$? How do I show $\arccos(4)$ is ${2\pi k}$ $\pm$ $i\operatorname{arcosh}(4)$?
I am getting $\pm$ $iln(4+\sqrt{15}$)
When I just use $\cos(z)$ =$\cosh(iz)$, I get:
$-i\operatorname{arcosh}(4)$.
I’m lost! If someone could type out the full proof, it'd be greatly appreciated. I'm not seeing it from the answers below. 
Thanks.
Edit: I follow the answers on the whole, but cannot seem to derive why arcosh has a prefix of $\pm$. If someone could show a derivation for this I'd be grateful. From function theory I'm expecting only the one positive root, not the -ve one. That said, I get the $\pm$ for the $\ln(x)$ version so I believe it should be there, but need a proof! :) Thanks. 
 A: $\arccos 4 = a+bi$
$4 = \cos(a+bi) = \cos a \cos bi - \sin a \sin bi = \cos a \cosh b - i\sin a \sinh b$
By comparing real and imaginary parts, 
$\cos a \cosh b = 4$ and $\sin a \sinh b = 0$
Where $a, b$ are both real. 
One possible solution to the second equation is $\sinh b =0 \implies b=0$, but that would make the first equation $\cos a = 4$, which has no real solution. So disregard this, i.e. $b \neq 0$.
Which leaves $\sin a =0 \implies a = n\pi, n \in \mathbb{Z}$.
In the first equation that makes $\pm \cosh b = 4$. Since the hyperbolic cosine for real values is non-negative only accept the positive case.
So $a = 2k\pi, k \in \mathbb{Z}$ (only even multiples of $\pi$ give a positive cosine) and $\cosh b = 4 \implies b = \pm \cosh^{-1} 4$ (since the hyperbolic cosine is an even function).
Thus we get the solution $\arccos 4 = 2k\pi \pm i\cosh^{-1} 4, k \in \mathbb{Z}$.
Note that you can also express the $\cosh^{-1} 4$ part in terms of logarithms if you wish. I believe this is what you did. To be clear $\cosh^{-1} 4 = \ln (4 + \sqrt {4^2 - 1}) = \ln (4 + \sqrt{15})$ , which means you can also write $\arccos 4 = 2k\pi \pm i \ln (4 + \sqrt{15}), k \in \mathbb{Z}$. This is an equivalent answer.
A: If we use $e^{i\theta}=\cos{(\theta)}+i\sin{(\theta)}$, we can show that $\cos{(\theta)}=\frac{e^{i\theta}+e^{-i\theta}}{2}$ or $\cos{(\theta)}=\cosh{(i\theta)}$.
$$
\begin{aligned}
\cosh{(\pm\cosh^{-1}{(4)})}&=4\\
\cosh{(i(2\pi n\pm i\cosh^{-1}{(4))})}&=4\\
\\
\cos{(2\pi n\pm i\cosh^{-1}{(4)})}&=4
\end{aligned}
$$
however... i do not know the range of $\cos^{-1}{(x)}$ in complex plane
A: The point is that 
$$\cosh(x +i 2 \pi k) = \cosh(x) = \cosh(-x)
$$
for $k$ integer.
A: Let $z = \arccos(4)$ and use the identity $\cos(iw)=\cosh(w)$ to express,
$$\cos(z) = \cos(2\pi n\pm z )
=\cos[i(-i2\pi n \mp iz )]=\cosh(-i2\pi n \mp iz ) = 4$$
or,
$$-i2n\pi \mp iz = \text{arccosh}(4)$$
which leads to,
$$z = \arccos(4) = 2\pi k \pm i\>\text{arccosh}(4) $$
