$\sinh$ and $\cosh$

So I recently found out that

$\sin {(a + bi)} = i\sinh {(b - ai)}$

$\cos {(a + bi)} = \cosh {(b - ai)}$

and that

$\sinh {(x)} = \frac {e^x - e^{-x}}{2}$

$\cosh {(x)} = \frac {e^x + e^{-x}}{2}$

Since

$\frac {a}{b} + \frac {b}{a} = \frac {a^2 + b^2}{ab}$

$\frac {a}{b} - \frac {b}{a} = \frac {a^2 - b^2}{ab}$

and

$(\frac {a}{b})^{-x} = (\frac {b}{a})^x$

Re-arranging the equations for $\sinh{(x)}$ and $\cosh {(x)}$ gives us

$\sinh {(x)} = \frac {e^{2x} - 1}{2e^x}$

$\cosh {(x)} = \frac {e^{2x} + 1}{2e^x}$

Substituting these into $\sin {(a + bi)}$ and $\cos {(a + bi)}$ when $x = b - ai$ gives us

$\sin {(a + bi)} = i\frac {e^{2(b - ai)} - 1}{2e^{b - ai}}$

$\cos {(a + bi)} = \frac {e^{2(b - ai)} + 1}{2e^{b - ai}}$

I'm going to try and expand and substitute by using

$e^{a + bi} = e^a (\cos {(b)} + i\sin {(b)})$

Let $\cos {\theta} + i\sin {\theta} =$ $cis$ $\theta$. So

$e^{a + bi} = e^a \cdot cis (b)$

Next we substitute this new equation in to get

$\sin {(a + bi)} = i\frac {e^{2b} \cdot cis (-2a) - 1}{2e^{b} \cdot cis (-a)}$

$\cos {(a + bi)} = \frac {e^{2b} \cdot cis (-2a) + 1}{2e^{b} \cdot cis (-a)}$

So what does $cis(a + bi)$ equal

Substituting in the two equations gives us

$cis(a + bi) = \frac {e^{2b} \cdot cis (-2a) + 1}{2e^{b} \cdot cis (-a)} + i(i\frac {e^{2b} \cdot cis (-2a) - 1}{2e^{b} \cdot cis (-a)})$

That looks unfriendly, let's simplify it

$cis(a + bi) = \frac {1}{e^{b} \cdot cis(-a)}$

(Go through the steps, it works)

This can be simplified even further

$cis(a + bi) = \frac {cis(a)}{e^b}$

P.s. $(\cos {\theta} + i\sin {\theta})^n = \cos {\theta n} + i\sin{\theta n}$