How does: $−i2nπ∓iz=arccosh(4) $ lead to, $z=arccos(4)=2πn±iarccosh(4)?$

Question

I'm confused by this step in a calculation:

$arcosh(4)=−i2nπ∓iz$, which leads to: $z=arccos(4)=2πn±iarcosh(4).$

From this I understand multiplying the LH expression by $i$ gives: $i.arcosh(4) = 2n\pi \pm z.$

But how do I move the $\pm$ in front of the $iarcosh(4)$? Some say it's because $cosh(x)$ is an even function, but it doesn't explain the $\pm$ for me. If someone could demonstrate this through a proof, it'd be very helpful! Thanks.

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Answer 1

Multiply $arcosh(4)=−i2nπ∓iz$ by $\pm 1$ and dividing by $i$ and writing $\pm 2n\pi $ as $2n\pi$ as $n$ being any integer will take care of $\pm.$ Just rearrange the expression as $z=...$