# How do you evaluate  e^( ( pi)/4 i) - e^( ( 35 pi)/4 i) using trigonometric functions?

Mar 28, 2016

${e}^{\frac{\pi}{4} i} - {e}^{\frac{35 \pi}{4} i} = \sqrt{2}$

#### Explanation:

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

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

${e}^{\frac{35 \pi}{4} i} = \cos \left(\frac{35 \pi}{4}\right) + i \sin \left(\frac{35 \pi}{4}\right)$

= $\cos \left(8 \pi + \frac{3 \pi}{4}\right) + i \sin \left(8 \pi + \frac{3 \pi}{4}\right)$

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

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

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

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

Hence ${e}^{\frac{\pi}{4} i} - {e}^{\frac{35 \pi}{4} i} = \left(\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}\right) - \left(- \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}\right)$

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

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