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

Jul 13, 2018

Euler's Formula states that

${e}^{i \textcolor{red}{\theta}} = \cos \textcolor{red}{\theta} + i \sin \textcolor{red}{\theta}$

Which is the trigonometric form of a complex number. Hence,

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

${e}^{i \textcolor{red}{\frac{7 \pi}{3}}} = \cos \left(\textcolor{red}{\frac{7 \pi}{3}}\right) + i \sin \left(\textcolor{red}{\frac{7 \pi}{3}}\right)$
$= \cos \left(2 \pi + \frac{\pi}{3}\right) + i \sin \left(2 \pi + \frac{\pi}{3}\right)$
$= \cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)$
$= \frac{1}{2} + i \frac{\sqrt{3}}{2}$

$\therefore {e}^{i \frac{\pi}{4}} - {e}^{i \frac{7 \pi}{3}} = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} - \frac{1}{2} - i \frac{\sqrt{3}}{2}$

$= \left(\frac{\sqrt{2}}{2} - \frac{1}{2}\right) + i \left(\frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2}\right)$