# How do you evaluate  e^( ( 7 pi)/4 i) - e^( ( 2 pi)/3 i) using trigonometric functions?

Jan 29, 2018

${e}^{\frac{7 \pi}{4} i} - {e}^{\frac{2 \pi}{3} i} = \frac{1}{2} + \frac{\sqrt{2}}{2} + i \left\{- \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2}\right\}$

#### Explanation:

For convenience, Let us denote the sum by:

$S = {e}^{\frac{7 \pi}{4} i} - {e}^{\frac{2 \pi}{3} i}$

We know from Euler's Formula , that:

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

And so we can represent the sum in trigonometric form by:

$S = \left(\cos \left(\frac{7 \pi}{4}\right) + i \sin \left(\frac{7 \pi}{4}\right)\right) - \left(\cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)\right)$

$\setminus \setminus = \cos \left(\frac{7 \pi}{4}\right) - \cos \left(\frac{2 \pi}{3}\right) + i \left\{\sin \left(\frac{7 \pi}{4}\right) - \sin \left(\frac{2 \pi}{3}\right)\right\}$

$\setminus \setminus = \frac{\sqrt{2}}{2} - \left(- \frac{1}{2}\right) + i \left\{- \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2}\right\}$

$\setminus \setminus = \frac{1}{2} + \frac{\sqrt{2}}{2} + i \left\{- \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2}\right\}$