Q:

Complex exponential to real

I'm not yet very good at complex number, so I would appreciate the following insight:
How exactly do we arrive from $e^{\pi(1-i)}-e^{-\pi(1-i)}$ to $e^{-π}-e^π$, and why does $e^{\pi(1-i)n}-e^{-\pi(1-i)n}$ have different signs for $n$ odd and $n$ even?

A:

For your first question, you must know by now that $e^{i\pi}=-1$, if you do not, then note that it follows from Euler's identity, since 
$$
e^{i\pi}=\cos(\pi)+i\sin(\pi) = -1+i0=-1 \tag{1}
$$
Hence 
\begin{align}
e^{\pi(1-i)}-e^{-\pi(1-i)} & = e^{\pi}e^{-\pi i}-e^{-\pi}e^{\pi i} \\ 
& = \frac{e^{\pi}}{e^{\pi i}}-e^{-\pi}e^{\pi i} \\ 
& \overset{(1)}{=} -e^\pi+e^{-\pi} \\
&= e^{-\pi} -e^\pi
\end{align}
For the second one, a similar line of thought shows that 
$$
e^{\pi(1-i)n}-e^{-\pi(1-i)n} = e^{\pi n}(-1)^n - e^{-\pi n}(-1)^n 
$$
Thus indeed the sign varies depending if $n$ is odd or even.

