Starting from the preceding line:
$\displaystyle \dfrac{1}{2\pi}\left(\dfrac{\left(-1\right)^n\left(e^{\pi} - e^{-\pi}\right)}{1-nj}\right)$
Move $\displaystyle \left(e^{\pi} - e^{-\pi}\right)$ to the numerator of $\displaystyle \dfrac{1}{2\pi}$ (commutative property of multiplication).
$\displaystyle \dfrac{\left(e^{\pi} - e^{-\pi}\right)}{2\pi}\left(\dfrac{\left(-1\right)^n}{1-nj}\right) = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{\left(-1\right)^n}{1-nj}$
Move $\displaystyle (-1)^n$ to the far right of the expression (commutative property of multiplication).
$\displaystyle \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{\left(-1\right)^n}{1-nj} = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1}{1-nj}\left(-1\right)^n$
Multiply by $\displaystyle \dfrac{1+nj}{1+nj}$.
$\displaystyle \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1}{1-nj}\left(-1\right)^n = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1}{1-nj}\dfrac{1+nj}{1+nj}\left(-1\right)^n = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1 + nj}{1 - n^2j^2}\left(-1\right)^n$
I assume $\displaystyle j = i = \sqrt{-1}$, in which case $\displaystyle j^2 = i^2 = -1$.
$\displaystyle \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1 + nj}{1 - n^2j^2}\left(-1\right)^n = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1 + nj}{1 - n^2\left(-1\right)}\left(-1\right)^n = \dfrac{e^{\pi} - e^{-\pi}}{2\pi}\dfrac{1 + nj}{1 + n^2}\left(-1\right)^n$
That's the way you do to find an expression of the form $\displaystyle a+bi$ for a fraction of the
form $\displaystyle \frac{1}{\alpha+i\beta}\,,\,\,a,b,\alpha,\beta\in\ mathbb{R}$.
The only weird thing is why you people use the letter "j" to denote what the huge majority
of mathematicians in the world denote by "i" = the imaginary unit, though I found that
in a very old book...
Tonio
All that is happening is that the top and bottom of the right hand side of the previous line are multiplied by $\displaystyle (1+nj)$ which is the complex conjugate of the denominator of the previous line. This renders the denominator real and moves all the complex terms into the numerator. Other than that there is some minor rearrangement of the other terms in the previous expression.
CB