# How do you multiply e^(( 4 pi )/ 3 i) * e^( 3 pi/2 i )  in trigonometric form?

Nov 18, 2017

${e}^{\frac{4 \pi}{3} i} \cdot {e}^{\frac{3 \pi}{2} i} = \cos \left(\frac{17 \pi}{2}\right) + i \sin \left(\frac{17 \pi}{2}\right)$

#### Explanation:

We can write exponential fom of complex number ${e}^{i \theta}$ in trigonometric form as $\cos \theta + i \sin \theta$

Hence ${e}^{\frac{4 \pi}{3} i} = \cos \left(\frac{4 \pi}{3}\right) + i \sin \left(\frac{4 \pi}{3}\right)$

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

Hence ${e}^{\frac{4 \pi}{3} i} \cdot {e}^{\frac{3 \pi}{2} i}$

= $\left(\cos \left(\frac{4 \pi}{3}\right) + i \sin \left(\frac{4 \pi}{3}\right)\right) \cdot \left(\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right)$

= $\cos \left(\frac{4 \pi}{3}\right) \cos \left(\frac{3 \pi}{2}\right) + {i}^{2} \sin \left(\frac{4 \pi}{3}\right) \sin \left(\frac{3 \pi}{2}\right) + i \left\{\sin \left(\frac{4 \pi}{3}\right) \cos \left(\frac{3 \pi}{2}\right) + \cos \left(\frac{4 \pi}{3}\right) \sin \left(\frac{3 \pi}{2}\right)\right\}$

= $\left\{\cos \left(\frac{4 \pi}{3}\right) \cos \left(\frac{3 \pi}{2}\right) - \sin \left(\frac{4 \pi}{3}\right) \sin \left(\frac{3 \pi}{2}\right)\right\} + i \left\{\sin \left(\frac{4 \pi}{3}\right) \cos \left(\frac{3 \pi}{2}\right) + \cos \left(\frac{4 \pi}{3}\right) \sin \left(\frac{3 \pi}{2}\right)\right\}$

= $\cos \left(\frac{4 \pi}{3} + \frac{3 \pi}{2}\right) + i \sin \left(\frac{4 \pi}{3} + \frac{3 \pi}{2}\right)$

= $\cos \left(\frac{17 \pi}{2}\right) + i \sin \left(\frac{17 \pi}{2}\right)$

Observe that this is ${e}^{\frac{17 \pi}{2}}$ which is nothing but ${e}^{\frac{4 \pi}{3} + \frac{3 \pi}{2}}$.