How do you multiply e^(( 9 pi )/ 4 i) * e^( pi/2 i )  in trigonometric form?

Sep 26, 2016

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

${e}^{\frac{9 \pi}{4} i} = \cos \left(\frac{9 \pi}{4}\right) + i \sin \left(\frac{9 \pi}{4}\right)$ and

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

Hence ${e}^{\frac{9 \pi}{4} i} \cdot {e}^{\frac{\pi}{2} i} = \left(\cos \left(\frac{9 \pi}{4}\right) + i \sin \left(\frac{9 \pi}{4}\right)\right) \left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right)$

= cos((9pi)/4)(cos(pi/2)+isin(pi/2))+isin((9pi)/4))(cos(pi/2)+isin(pi/2))

= cos((9pi)/4)cos(pi/2)+icos((9pi)/4)sin(pi/2))+isin((9pi)/4)cos(pi/2)+i^2sin((9pi)/4)sin(pi/2))

= cos((9pi)/4)cos(pi/2)+icos((9pi)/4)sin(pi/2))+isin((9pi)/4)cos(pi/2)-sin((9pi)/4)sin(pi/2))

= $\left(\cos \left(\frac{9 \pi}{4}\right) \cos \left(\frac{\pi}{2}\right) - \sin \left(\frac{9 \pi}{4}\right) \sin \left(\frac{\pi}{2}\right)\right) + i \left(\sin \left(\frac{9 \pi}{4}\right) \cos \left(\frac{\pi}{2}\right) + \cos \left(\frac{9 \pi}{4}\right) \sin \left(\frac{\pi}{2}\right)\right)$

= $\cos \left(\left(\frac{9 \pi}{4}\right) + \left(\frac{\pi}{2}\right)\right) + i \sin \left(\left(\frac{9 \pi}{4}\right) + \left(\frac{\pi}{2}\right)\right)$

= $\cos \left(\frac{11 \pi}{4}\right) + i \sin \left(\frac{11 \pi}{4}\right)$

= ${e}^{\frac{11 \pi}{4}}$