so by using the identities we have:

$4=\cos (a)\cosh (b)-i\sin (a)\sinh (b)$

so by equating real and imaginary parts we have:

(1) $\cos (a)\cosh (b)=4$ and (2) $\sin (a)\sinh (b)=0$

because $b$ form $a+ib$ on the left hand side is zero and $a$ is 4. From (2) we can see that either $\sin (a)$ is zero $\Rightarrow a=0$ or $\sinh (b)=0\Rightarrow b=0$. So let's assume that $b$ is zero, so from (1) we have:

$\cos (a)\cosh (0)=4\Rightarrow \cos (a)=4$

Now we have a problem because $a$ is a real number and $\cos (a)$ is not greater than 1 for real $a$, so $b$ cannot be zero and $a$ is zero:

$\cos (0)\cosh (b)=4\Rightarrow \cosh (b)=4\Rightarrow b=\pm \mathrm{arcosh}(4)$

The $\pm$ because cosh is an even function.

But $a$ can also be $2k\pi$ for any integer $k$ because $\cos (2k\pi )$ is still 1 and $\sin (2k\pi )$ is still 0. So:

$\arccos (4)=2k\pi \pm i\mathrm{arcosh}(4)$.