# How do you multiply e^(( 3 pi )/ 8 i) * e^( 3 pi/2 i )  in trigonometric form?

Oct 15, 2017

$= {e}^{\left(\frac{31 \pi}{8}\right) i}$

#### Explanation:

 e^(((3pi)/8)i) * e^ (3(pi/2)i
${e}^{\left(\frac{3 \pi}{8}\right) i} \cdot {e}^{\left(\frac{7 \pi}{2}\right) i}$

${e}^{\left(\frac{3 \pi}{8}\right) i} \cdot {e}^{\left(\frac{28 \pi}{8}\right) i}$
e^ (((3pi + 28pi)/8)(i+i) = e^(((31pi)/8)i)

Oct 15, 2017

$\cos \left(\frac{\pi}{8}\right) - i \sin \left(\frac{\pi}{8}\right)$ or $- {\left(- 1\right)}^{\frac{7}{8}}$

#### Explanation:

Are you sure its ${e}^{3 \frac{\pi}{2}}$ not ${e}^{\frac{3 \pi}{2}}$ because it makes more sense when you write it in trig form.

Remember this beauty?
${e}^{i \pi} = - 1$
That was Euler's identity. This is the generalized formula:
${e}^{i x} = \cos x + i \sin x$

Therefore, we can break down the two terms involved:

${e}^{i \frac{3}{8} \pi} = \cos \left(\frac{3 \pi}{8}\right) + i \sin \left(\frac{3 \pi}{8}\right)$

${e}^{i \frac{7}{2} \pi} = \cos \left(\frac{7 \pi}{2}\right) + i \sin \left(\frac{7 \pi}{2}\right) = 0 + i \cdot \left(- 1\right) = - i$

${e}^{i \frac{3}{8} \pi} = {\left({e}^{i \frac{3}{2} \pi}\right)}^{\frac{1}{4}} = {\left(- i\right)}^{\frac{1}{4}}$
${e}^{i \frac{3}{8} \pi} \cdot {e}^{i \frac{3}{2} \pi} = {\left({e}^{i \frac{3}{2} \pi}\right)}^{\frac{1}{4}} \cdot {e}^{i \frac{7}{2} \pi} = {\left({e}^{i \frac{3}{2} \pi}\right)}^{\frac{5}{4}} = {\left(- i\right)}^{\frac{5}{4}} = - {\left(- 1\right)}^{\frac{7}{8}}$

Or

${e}^{i \frac{3}{8} \pi} \cdot {e}^{i \frac{7}{2} \pi} = \sin \left(\frac{3 \pi}{8}\right) - i \cos \left(\frac{3 \pi}{8}\right)$

$= \cos \left(\frac{\pi}{8}\right) - i \sin \left(\frac{\pi}{8}\right)$