Given the following expression
$$w = j \cos \left[ \displaystyle \frac{1}{n} \arccos \left( \frac{j}{\epsilon} \right) + \frac{m \pi}{n} \right] = j \cos (z)$$
(which is related to this question; $n,m \in \mathbf{Z}$, $\epsilon > 0 \in \mathbf{R}$), first of all I would like to express the real and imaginary part of $w$ respectively in terms of $\sinh$ and $\cosh$ of some appropriate angles. The aim is to obtain something like
$$j \cos(z) = A \sinh(u) + j B \cosh(v)$$
1) Is it possible?
I tried
$$j \cos (z) = j \frac{e^{jz} + e^{-jz}}{2} = \frac{e^{j\left( z + \frac{\pi}{2} \right)} + e^{- j\left( z - \frac{\pi}{2} \right)}}{2}$$
but up to now I don't see a connection with $\sinh$ and $\cosh$. Nevertheless:
2) if this could be a suitable way, how to proceed? Or are there other alternative ways?
Set $z=x+iy$, where $x,y$ are real. Use the addition formula: $$ \cos{(x+iy)} = \cos{x}\cos{iy}-\sin{x}\sin{iy}. $$ Now you have $$ \cos{iy} = \cosh{y}, \quad \sin{iy} = i\sinh{y}, $$ which is easy to prove in the way you suggested. So $$ i\cos{(x+iy)} = i(\cos{x}\cosh{y}-i\sin{x}\sinh{y}) = i\cos{x}\cosh{y}+\sin{x}\sinh{y}. $$