Question: A question in my textbook asks:

Find $\frac{z\_1}{z\_2}$ if $z\_1=2\left(\cos\left(\frac{\pi}3\right)+i\sin\left(\frac{\pi}3\right)\right)$ and $z\_2=\cos\left(\frac{\pi}6\right)-i\sin\left(\frac{\pi}6\right)$. I converted $z\_2$ to $\cos\left(-\frac{\pi}6\right)+i\sin\left(-\frac{\pi}6\right)$ as I initially thought it would be easier to use Euler's identity (which it is) but the textbook hadn't introduced this yet so it must be possible without having to use it. As a result, I am stuck at square one, any help would be great. Thanks.

Answer: 
If you are working with complex number in the form you gave, recall that $r\cos\theta+ir\sin\theta=re^{i\theta}$.

You then multiply and divide complex numbers in polar form in the natural way:

$$r\_1e^{1\theta\_1}\cdot r\_2e^{1\theta\_2}=r\_1r\_2e^{i(\theta\_1+\theta\_2)},$$

$$\frac{r\_1e^{1\theta\_1}}{r\_2e^{1\theta\_2}}=\frac{r\_1}{r\_2}e^{i(\theta\_1-\theta\_2)}$$
$$z\_{1}=2(cos(\frac{pi}{3})+i sin (\frac{pi}{3}) )=2e^{i\frac{pi}{3}}\\z\_{2}=1(cos(\frac{pi}{6})-i sin (\frac{pi}{6}) )=1(cos(\frac{pi}{6})
+i sin (\frac{-pi}{6}) )=\\as-we-know\\cos(a)=cos(-a)\\1(cos(\frac{-pi}{6})-i sin (\frac{-pi}{6}) )=1e^{\frac{-pi}{6}\\
$$