How do you evaluate  e^( ( 5 pi)/4 i) - e^( ( 3 pi)/4 i) using trigonometric functions?

Jan 28, 2018

$\sqrt{2} {e}^{\frac{3 \pi}{2} \cdot i}$

Explanation:

${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$ so:

${e}^{\frac{5 \pi}{4} i} = \cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right) = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i$
and
${e}^{\frac{3 \pi}{4} i} = \cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right) = - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i$

Which means that

${e}^{\frac{5 \pi}{4} i} - {e}^{\frac{3 \pi}{4} i} = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i - \left(- \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i\right)$

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

$= - \sqrt{2} i$

which is a complex number with $r = \sqrt{2}$ and $\theta = \frac{3 \pi}{2}$

so

${e}^{\frac{5 \pi}{4} i} - {e}^{\frac{3 \pi}{4} i} = \sqrt{2} {e}^{\frac{3 \pi}{2} \cdot i}$