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

Sep 30, 2016

${e}^{\frac{\pi}{2} i} \cdot {e}^{\pi i} = {e}^{\frac{\pi}{2} + \pi}$

#### Explanation:

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

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

${e}^{\pi i} = \cos \pi + i \sin \pi$

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

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

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

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

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

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

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