# How do you evaluate  e^( (3 pi)/2 i) - e^( ( pi)/3 i) using trigonometric functions?

Feb 26, 2017

Use Euler's identity:

${e}^{i x} = \cos x + i \sin x$

Thus:

${e}^{\frac{3 \pi}{2} i} - {e}^{\frac{\pi}{3} i}$

$= \cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right) - \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)$

$= {\cancel{\cos \left(\frac{3 \pi}{2}\right)}}^{0} + i \sin \left(\frac{3 \pi}{2}\right) - \cos \left(\frac{\pi}{3}\right) - i \sin \left(\frac{\pi}{3}\right)$

$= - i - \frac{1}{2} - \frac{\sqrt{3}}{2} i$

$= \textcolor{b l u e}{- \frac{1}{2} + \left(- 1 - \frac{\sqrt{3}}{2}\right) i}$

So, for the form $a + b i$, $a = - \frac{1}{2}$ and $b = - 1 - \frac{\sqrt{3}}{2}$.