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

Jan 22, 2017

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

#### Explanation:

Euler's formula states

${e}^{i x} = \cos \left(x\right) + i \sin \left(x\right)$

Then for $x = \frac{\pi}{2}$

${e}^{i x} = {e}^{x i} = {e}^{\frac{\pi}{2} i} = \cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right) = 0 + i \left(1\right) = i$

and for $x = \frac{7 \pi}{4} = \left(2 \pi - \frac{\pi}{4}\right) \implies - \frac{\pi}{4}$

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

Then plug in

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

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