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

Mar 27, 2016

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

#### Explanation:

As ${e}^{i \theta} = \cos \theta + i \sin \theta$, we have

${e}^{\frac{\pi}{4} i} = \cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)$ and

${e}^{\frac{7 \pi}{4} i} = \cos \left(\frac{7 \pi}{4}\right) + i \sin \left(\frac{7 \pi}{4}\right)$

Hence, ${e}^{\frac{\pi}{4} i} - {e}^{\frac{7 \pi}{4} i}$

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

As $\cos \left(\frac{\pi}{4}\right) = \cos \left(\frac{7 \pi}{4}\right) = \frac{1}{\sqrt{2}}$,

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

${e}^{\frac{\pi}{4} i} - {e}^{\frac{7 \pi}{4} i}$

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

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

=$0 + i \frac{2}{\sqrt{2}}$ = $\sqrt{2} i$