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

Sep 30, 2016

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

${e}^{\frac{\pi}{8} i} = \cos \left(\frac{\pi}{8}\right) + i \sin \left(\frac{\pi}{8}\right)$ and

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

and ${e}^{\frac{\pi}{8} i} \cdot {e}^{\frac{\pi}{2} i} = \left(\cos \left(\frac{\pi}{8}\right) + i \sin \left(\frac{\pi}{8}\right)\right) \left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right)$

= $\cos \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{2}\right) + \cos \left(\frac{\pi}{8}\right) \times i \sin \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{8}\right) \times \cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{8}\right) \times i \sin \left(\frac{\pi}{2}\right)$

= $\cos \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{2}\right) + i \cos \left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{2}\right) + {i}^{2} \sin \left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{2}\right)$

= $\cos \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{2}\right) + i \cos \left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{2}\right) - \sin \left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{2}\right)$

= $\left[\cos \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{2}\right) - \sin \left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{2}\right)\right] + i \left[\cos \left(\frac{\pi}{2}\right) \sin \left(\frac{\pi}{8}\right) + \sin \left(\frac{\pi}{2}\right) \cos \left(\frac{\pi}{8}\right)\right]$

= $\cos \left(\frac{\pi}{8} + \frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{8} + \frac{\pi}{2}\right)$

= ${e}^{\frac{\pi}{8} + \frac{\pi}{2}}$