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

Aug 6, 2017

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

#### Explanation:

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

Here, we have ${e}^{\frac{7 \pi}{4} i} - {e}^{\frac{4 \pi}{3} i} = \cos \left(\frac{7 \pi}{4}\right) + i \sin \left(\frac{7 \pi}{4}\right) - \left(\cos \left(\frac{4 \pi}{3}\right) + i \sin \left(\frac{4 \pi}{3}\right)\right) = \cos \left(\frac{7 \pi}{4}\right) + i \sin \left(\frac{7 \pi}{4}\right) - \cos \left(\frac{4 \pi}{3}\right) - i \sin \left(\frac{4 \pi}{3}\right)$

$\cos \left(\frac{7 \pi}{4}\right) = \frac{\sqrt{2}}{2}$
$\sin \left(\frac{7 \pi}{4}\right) = - \frac{\sqrt{2}}{2}$
$\cos \left(\frac{4 \pi}{3}\right) = - \frac{1}{2}$
$\sin \left(\frac{4 \pi}{3}\right) = - \frac{\sqrt{3}}{2}$

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