# How do you simplify 3(cos(pi/6)+isin(pi/6))div4(cos((2pi)/3)+isin((2pi)/3)) and express the result in rectangular form?

Dec 14, 2016

The answer is $= - \frac{3}{4} i$

#### Explanation:

There are 2 ways for the simplification

$\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$

$\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

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

$\sin \left(2 \frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

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

$\frac{3 \left(\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)\right)}{4 \left(\cos \left(2 \frac{\pi}{3}\right) + i \sin \left(2 \frac{\pi}{3}\right)\right)}$

$= \frac{3}{4} \frac{\frac{\sqrt{3}}{2} + i \cdot \frac{1}{2}}{- \frac{1}{2} + i \frac{\sqrt{3}}{2}}$

$= \frac{3}{4} \frac{\left(\frac{\sqrt{3}}{2} + i \cdot \frac{1}{2}\right) \left(- \frac{1}{2} - i \frac{\sqrt{3}}{2}\right)}{\left(- \frac{1}{2} + i \frac{\sqrt{3}}{2}\right) \left(- \frac{1}{2} - i \frac{\sqrt{3}}{2}\right)}$

$= \frac{3}{4} \frac{- \frac{\sqrt{3}}{4} - i \frac{3}{4} - \frac{i}{4} + \frac{\sqrt{3}}{4}}{\frac{1}{4} + \frac{3}{4}}$

$= \frac{3}{4} \left(- i\right)$

We can also use $\cos \theta + i \sin \theta = {e}^{i \theta}$

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

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

$\therefore \frac{3 \left(\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)\right)}{4 \left(\cos \left(2 \frac{\pi}{3}\right) + i \sin \left(2 \frac{\pi}{3}\right)\right)} = \frac{3}{4} {e}^{i \frac{\pi}{6}} / {e}^{2 i \frac{\pi}{3}}$

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

$= \frac{3}{4} {e}^{- i \frac{\pi}{2}}$

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

$= \frac{3}{4} \cdot \left(0 - i\right)$

$= - \frac{3 i}{4}$

Dec 14, 2016

$3 \left(\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)\right) \div 4 \left(\cos \left(2 \frac{\pi}{3}\right) + i \sin \left(2 \frac{\pi}{3}\right)\right) = - \frac{3}{4} i$

#### Explanation:

Given two complex numbers ${z}_{1} = {r}_{1} \left(\cos \alpha + i \sin \alpha\right)$ and ${z}_{2} = {r}_{2} \left(\cos \beta + i \sin \beta\right)$

${z}_{1} \div {z}_{2} = {r}_{1} / {r}_{2} \left(\cos \left(\alpha - \beta\right) + i \in \left(\alpha - \beta\right)\right)$

$3 \left(\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)\right) \div 4 \left(\cos \left(2 \frac{\pi}{3}\right) + i \sin \left(2 \frac{\pi}{3}\right)\right)$

= $\frac{3}{4} \left(\cos \left(\frac{\pi}{6} - \frac{2 \pi}{3}\right) + i \sin \left(\frac{\pi}{6} - \frac{2 \pi}{3}\right)\right)$

= $\frac{3}{4} \left(\cos \left(\frac{\pi}{6} - \frac{4 \pi}{6}\right) + i \sin \left(\frac{\pi}{6} - \frac{4 \pi}{6}\right)\right)$

= $\frac{3}{4} \left(\cos \left(- \frac{3 \pi}{6}\right) + i \sin \left(- \frac{3 \pi}{6}\right)\right)$

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

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

= $\frac{3}{4} \left(0 - i\right)$

= $- \frac{3}{4} i$