I understand the following Euler trigonometric identities:
$$sin(\omega t)=\frac{e^{j\omega t}-e^{-j\omega t}}{2j}$$
and
$$e^{j\omega t}-e^{-j\omega t} = 2jsin(\omega t)$$
What I dont understand is how $-3je^{j\omega t}+3je^{-j\omega t} =6\left [\frac{e^{j\omega t}-e^{-j\omega t}}{2j} \right ]$
This is regarding Fourier Series coefficients where $\pm 3J$ are the coefficients. I'm using the above concepts and the answer should be $6sin(\omega t)$, however the use of the complex number J is confusing me. Any assistance is appreciated.