Question #89196

Nov 24, 2016

${z}_{1} = 2 {e}^{\frac{3 \pi}{2} i}$
${z}_{2} = 8 {e}^{\frac{4 \pi}{3} i}$
${z}_{1} {z}_{2} = 16 {e}^{\frac{5 \pi}{6} i}$
${z}_{1} / {z}_{2} = \frac{1}{4} {e}^{\frac{\pi}{6} i}$

Explanation:

Using Euler's formula ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$.

${z}_{1} = - 2 i$

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

${z}_{2} = - 4 - 4 \sqrt{3} i$

$= 8 \left(- \frac{1}{2} - \frac{\sqrt{3}}{2} i\right)$
$= 8 \left(\cos \left(\frac{4 \pi}{3}\right) + i \sin \left(\frac{4 \pi}{3}\right)\right)$
$= 8 {e}^{\frac{4 \pi}{3} i}$

Next, using that ${e}^{x} {e}^{y} = {e}^{x y}$ and ${e}^{x} / {e}^{y} = {e}^{x - y}$, we have

${z}_{1} {z}_{2} = 2 {e}^{\frac{3 \pi}{2} i} \cdot 8 {e}^{\frac{4 \pi}{3} i} = 16 {e}^{\left(\frac{3}{2} + \frac{4}{3}\right) \pi i} = 16 {e}^{\frac{17 \pi}{6} i} = 16 {e}^{\frac{5 \pi}{6} i}$

where the last step uses ${e}^{i \theta} = {e}^{i \left(\theta + 2 \pi n\right)} , n \in \mathbb{Z}$

and

${z}_{1} / {z}_{2} = \frac{2 {e}^{\frac{3 \pi}{2} i}}{8 {e}^{\frac{4 \pi}{3} i}} = \frac{1}{4} {e}^{\left(\frac{3}{2} - \frac{4}{3}\right) \pi i} = \frac{1}{4} {e}^{\frac{\pi}{6} i}$