Separate real and imaginary part of $\arccos(z)$ Beginning with
$$i \cos \left[ \frac{1}{n} \arccos \left( \frac{i}{\epsilon} \right) + \frac{m \pi}{n} \right]$$
where $m,n \in \mathbf{Z}$, $\epsilon >0$, $\epsilon \in \mathbf{R}$ and $i$ is the imaginary unit, I would like to obtain separately the real and imaginary part of the cosine argument:
(1)
$$\frac{1}{n} \arccos \left( \frac{i}{\epsilon} \right) + \frac{m \pi}{n} = x + iy$$
By simply applying the definition:
(2)
$$\frac{1}{n} \arccos \left( \frac{i}{\epsilon} \right) = \frac{\pi}{2n} + i \frac{1}{n} \ln \left[ \frac{1}{\epsilon} (\sqrt{1 + \epsilon^2} - 1) \right]$$
Similarly, if we apply the definition of complex $\arcsin$, we obtain:
$$\arcsin \left( \frac{i}{\epsilon} \right) = -i \ln \left[ \frac{1}{\epsilon} (\sqrt{1 + \epsilon^2} - 1) \right]$$
Remembering (as stated here) that $$\mathrm{arsinh}(z) = -i\arcsin(iz)$$ we obtain
$$\mathrm{arsinh} \left( \frac{1}{\epsilon} \right) = - \ln \left[ \frac{1}{\epsilon} (\sqrt{1 + \epsilon^2} - 1) \right]$$
So we can write
$$x + iy = \frac{m \pi}{n} + \frac{\pi}{2n} + i\frac{1}{n} \ln \left[ \frac{1}{\epsilon} (\sqrt{1 + \epsilon^2} - 1) \right]$$
$$x + iy = \frac{\pi}{2} \left( \frac{2m + 1}{n} \right) - i\frac{1}{n} \mathrm{arsinh} \left( \frac{1}{\epsilon} \right) = \frac{\pi}{2} \left( \frac{2m - 1}{n} \right) - i\frac{1}{n} \mathrm{arsinh} \left( \frac{1}{\epsilon} \right)$$
But the correct result should be
(3)
$$x + iy = \frac{\pi}{2} \left( \frac{2m - 1}{n} \right) + i\frac{1}{n} \mathrm{arsinh} \left( \frac{1}{\epsilon} \right)$$
So, are there any errors? How to obtain (3) from (1)?

$$i \cos \left( \frac{1}{n} \cos^{-1} \left( \frac{i}{\epsilon} \right) + \frac{m \pi}{n} \right)=i\cos\left(\frac{\cos^{-1}\left(\frac{i}{\epsilon}\right)+m\pi}{n}\right)=$$
$$i\cos\left(\frac{\sec^{-1}\left(-i\epsilon\right)+m\pi}{n}\right)=i\cos\left(\frac{\frac{\pi}{2}-i\sinh^{-1}\left(\frac{1}{\epsilon}\right)}{n}+ \frac{m\pi}{n}\right)=$$
$$i\cos\left(\frac{\pi}{2n}-\frac{i\sinh^{-1}\left(\frac{1}{\epsilon}\right)}{n}+ \frac{m\pi}{n}\right)=$$
$$i\cos\left(\frac{\pi+2m\pi}{2n}-\frac{i*csch^{-1}(\epsilon)}{n}\right)=$$
$$i\cos\left(\frac{\pi+2m\pi}{2n}-\frac{csch^{-1}(\epsilon)}{n}i\right)$$

$$\frac{\pi+2m\pi}{2n}-\frac{csch^{-1}(\epsilon)}{n}i=\frac{\pi}{2}\left(\frac{2m-1}{n}\right)-\frac{\sinh^{-1}\left(\frac{1}{\epsilon}\right)}{n}i=\frac{\pi}{2}\left(\frac{2m-1}{n}\right)-i\frac{1}{n}\sinh^{-1}\left(\frac{1}{\epsilon}\right)$$
