# How do you multiply e^(( 3 pi )/ 2 i) * e^( 3 pi/2 i )  in trigonometric form?

Mar 27, 2016

${e}^{\frac{3 \pi}{2} i} \cdot {e}^{\frac{3 \pi}{2} i} = - 1$

#### Explanation:

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

${e}^{\frac{3 \pi}{2} i} = \cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)$.

Hence, ${e}^{\frac{3 \pi}{2} i} \cdot {e}^{\frac{3 \pi}{2} i} =$

$\left\{\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right\} \cdot \left\{\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right\}$ or

= $\left\{{\cos}^{2} \left(\frac{3 \pi}{2}\right) + {i}^{2} {\sin}^{2} \left(\frac{3 \pi}{2}\right) + 2 i \sin \left(\frac{3 \pi}{2}\right) \cos \left(\frac{3 \pi}{2}\right)\right\}$

= $\left\{{\cos}^{2} \left(\frac{3 \pi}{2}\right) + \left(- 1\right) {\sin}^{2} \left(\frac{3 \pi}{2}\right) + i \left(2 \sin \left(\frac{3 \pi}{2}\right) \cos \left(\frac{3 \pi}{2}\right)\right)\right\}$

= $\left\{\left({\cos}^{2} \left(\frac{3 \pi}{2}\right) - {\sin}^{2} \left(\frac{3 \pi}{2}\right)\right) + i \left(2 \sin \left(\frac{3 \pi}{2}\right) \cos \left(\frac{3 \pi}{2}\right)\right)\right\}$

= $\cos \left(2 \times \frac{3 \pi}{2}\right) + i \sin \left(2 \times \frac{3 \pi}{2}\right)$

= $\cos 3 \pi + i \sin 3 \pi$

= $- 1 + i \times 0 = - 1$