# How do you multiply e^((11pi )/ 12 ) * e^( pi/4 i )  in trigonometric form?

May 30, 2018

The answer is $= - \frac{\sqrt{3}}{2} - \frac{1}{2} i$

#### Explanation:

Apply Euler's relation

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

${e}^{\frac{11}{12} i \pi} = \cos \left(\frac{11}{12} \pi\right) + i \sin \left(\frac{11}{12} \pi\right)$

${e}^{\frac{1}{4} i \pi} = \cos \left(\frac{1}{4} \pi\right) + i \sin \left(\frac{1}{4} \pi\right)$

${i}^{2} = - 1$

Therefore,

${e}^{\frac{11}{12} i \pi} \cdot {e}^{\frac{1}{4} i \pi} = \left(\cos \left(\frac{11}{12} \pi\right) + i \sin \left(\frac{11}{12} \pi\right)\right) \cdot \left(\cos \left(\frac{1}{4} \pi\right) + i \sin \left(\frac{1}{4} \pi\right)\right)$

=cos(11/12pi)cos(1/4pi)-sin(11/12pi)sin(1/4pi)+i(cos(11/12pi)sin(1/4pi)+sin(11/12pi)cos(1/4pi)))

$= \cos \left(\frac{11}{12} \pi + \frac{1}{4} \pi\right) + i \sin \left(\frac{11}{12} \pi + \frac{1}{4} \pi\right)$

$= \cos \left(\frac{14}{12} \pi\right) + i \sin \left(\frac{14}{12} \pi\right)$

$= \cos \left(\frac{7}{6} \pi\right) + i \sin \left(\frac{7}{6} \pi\right)$

$= - \frac{\sqrt{3}}{2} - \frac{1}{2} i$