Fourier Series coefficients - How does one separate e^(jwt) from the rest of the answer?

Question

How does one go from the first step to the second step? (In red squares in attachment here)

I don't understand how to separate the complex form or separate the e^(jwt) from the result. Could someone explain? Thanks!

Answer 1

$$ \mathrm{e}^{\mathrm{i}\pi/6} = \cos(\pi/6) + \mathrm{i}\,\sin(\pi/6) = \frac{\sqrt{3}}{2} + \frac{\mathrm{i}}{2} \text{.} $$ $$ \mathrm{e}^{-\mathrm{i}\pi/3} = \cos(-\pi/3) + \mathrm{i}\,\sin(-\pi/3) = \frac{-1}{2} + \frac{-\mathrm{i}\sqrt{3}}{2} \text{.} $$ (Your source has an error on the sign of the real part. It's duplicated in the fourth term.) $$ 2 = 2 \text{.} $$ $$ \mathrm{e}^{\mathrm{i}\pi/3} = \cos(\pi/3) + \mathrm{i}\,\sin(\pi/3) = \frac{-1}{2} + \frac{\mathrm{i}\sqrt{3}}{2} \text{.} $$ $$ \mathrm{e}^{-\mathrm{i}\pi/6} = \cos(-\pi/6) + \mathrm{i}\,\sin(-\pi/6) = \frac{\sqrt{3}}{2} + \frac{-\mathrm{i}}{2} \text{.} $$

These are all specializations of Euler's formula, $$ \mathrm{e}^{\mathrm{i}\theta} = \cos(\theta) +\mathrm{i} \,\sin(\theta) \text{.} $$