I need to prove
$$\frac{e^{(-1)^{\frac{2}{\alpha}}x}-e^{-(-1)^{\frac{2}{\alpha}}x}}{2i}=e^{-x\cos(\frac{2\pi}{\alpha}))}\sin(x\sin(\frac{2\pi}{\alpha}))$$
My attempt:
we can write $$e^{(-1)^{\frac{2}{\alpha}}x}=\exp\{x\cos(\frac{2\pi}{\alpha})+ix\sin(\frac{2\pi}{\alpha})\}$$
Using this I can get $$\frac{\exp\{x\cos(\frac{2\pi}{\alpha})\}\exp\{ix\sin(\frac{2\pi}{\alpha})\}-\exp\{-x\cos(\frac{2\pi}{\alpha})\}\exp\{-ix\sin(\frac{2\pi}{\alpha})\}}{2i}$$ but this expression is not equal to the right side of my first equation. Am I doing something terribly wrong? BTW the expression on the right side of my first equation is from an IEEE transactions paper.
It is not correct. Try $\alpha=2$ to get
$$\frac{e^{-x}-e^x}{2i}=e^x\sin(x.0)=0$$