# How do you multiply e^(( 5 pi )/ 4 i) * e^( 3 pi/2 i )  in trigonometric form?

Apr 11, 2018

Using the formula ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$ and some trigonometric identities.

#### Explanation:

The formula:

${e}^{\frac{5 \pi}{4} i} = \cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right)$

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

Multiplication:

${e}^{\frac{5 \pi}{4} i} \times {e}^{\frac{3 \pi}{2} i} = \left(\cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right)\right) \left(\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right)$

$= \cos \left(\frac{5 \pi}{4}\right) \cos \left(\frac{3 \pi}{2}\right) + i \cos \left(\frac{5 \pi}{4}\right) \sin \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{5 \pi}{4}\right) \cos \left(\frac{3 \pi}{2}\right) - \sin \left(\frac{5 \pi}{4}\right) \sin \left(\frac{3 \pi}{2}\right)$

$= \cos \left(\frac{5 \pi}{4}\right) \cos \left(\frac{3 \pi}{2}\right) - \sin \left(\frac{5 \pi}{4}\right) \sin \left(\frac{3 \pi}{2}\right) + i \left(\cos \left(\frac{5 \pi}{4}\right) \sin \left(\frac{3 \pi}{2}\right) + \sin \left(\frac{5 \pi}{4}\right) \cos \left(\frac{3 \pi}{2}\right)\right)$

$= \cos \left(\frac{5 \pi}{4} + \frac{3 \pi}{2}\right) + i \sin \left(\frac{5 \pi}{4} + \frac{3 \pi}{2}\right)$

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

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

Its much easier if you do this:

${e}^{\frac{5 \pi}{4} i} \times {e}^{\frac{3 \pi}{2} i} = {e}^{\frac{5 \pi}{4} i + \frac{3 \pi}{2} i}$

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

Apr 11, 2018

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

#### Explanation:

Apply Eulers' Identity

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

Therefore,

${e}^{\frac{5}{4} \pi i} \cdot {e}^{\frac{3}{2} \pi i} = \left(\cos \left(\frac{5}{4} \pi\right) + i \sin \left(\frac{5}{4} \pi\right)\right) \left(\cos \left(\frac{3}{2} \pi\right) + i \sin \left(\frac{3}{2} \pi\right)\right)$

$\cos \left(\frac{5}{4} \pi\right) = \cos \left(\frac{1}{4} \pi + \pi\right) = \cos \left(\frac{1}{4} \pi\right) \cos \left(\pi\right) - \sin \left(\frac{1}{4} \pi\right) \sin \left(\pi\right)$

$= - \frac{\sqrt{2}}{2}$

$\sin \left(\frac{5}{4} \pi\right) = \sin \left(\frac{1}{4} \pi + \pi\right) = \sin \left(\frac{1}{4} \pi\right) \cos \left(\pi\right) + \cos \left(\frac{1}{4} \pi\right) \sin \left(\pi\right)$

$= - \frac{\sqrt{2}}{2}$

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

$= 0$

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

$= - 1$

Finally,

${e}^{\frac{5}{4} \pi i} \cdot {e}^{\frac{3}{2} \pi i} = \left(- \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i\right) \left(- i\right)$

$= - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i$