How do you multiply e^(( 9 pi )/ 8 i) * e^( pi/2 i )  in trigonometric form?

Dec 30, 2016

${e}^{\frac{9 \pi}{8} i} \cdot {e}^{\frac{\pi}{2} i} = \textcolor{g r e e n}{\sin \left(\frac{\pi}{8}\right) - i \cos \left(\frac{\pi}{8}\right)}$

Explanation:

Using Euler's formula
$\textcolor{w h i t e}{\text{XXX}} {e}^{\frac{9 \pi}{8} i} = \cos \left(\frac{9 \pi}{8}\right) + i \cdot \sin \left(\frac{9 \pi}{8}\right)$

$\textcolor{w h i t e}{\text{XXXXX}} = \cos \left(\pi + \frac{\pi}{8}\right) + i \cdot \sin \left(\pi + \frac{\pi}{8}\right)$

$\textcolor{w h i t e}{\text{XXXXX}} = - \cos \left(\frac{\pi}{8}\right) + i \cdot \left(- \sin \left(\frac{\pi}{8}\right)\right)$

$\textcolor{w h i t e}{\text{XXXXX}} = - \cos \left(\frac{\pi}{8}\right) - i \cdot \sin \left(\frac{\pi}{8}\right)$

and
$\textcolor{w h i t e}{\text{XXX}} {e}^{\frac{\pi}{2} i} = \cos \left(\frac{\pi}{2}\right) + i \cdot \sin \left(\frac{\pi}{2}\right)$

$\textcolor{w h i t e}{\text{XXXXX}} = 0 + i \cdot 1$

$\textcolor{w h i t e}{\text{XXXXX}} = i$

Therefore
$\textcolor{w h i t e}{\text{XXX}} {e}^{\frac{9 \pi}{8} i} \cdot {e}^{\frac{\pi}{2} i} = \left[- \cos \left(\frac{\pi}{8}\right) - i \cdot \sin \left(\frac{\pi}{8}\right)\right] \cdot i$

$\textcolor{w h i t e}{\text{XXXXXXX}} = - i \cdot \cos \left(\frac{\pi}{8}\right) - \left(- 1\right) \cdot \sin \left(\frac{\pi}{8}\right)$

$\textcolor{w h i t e}{\text{XXXXXXX}} = \sin \left(\frac{\pi}{8}\right) - i \cos \left(\frac{\pi}{8}\right)$