How do you divide complex numbers in polar form? 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.
Just an expansion of my comment above: presumably you know how to do
$$
\alpha(a+bi)(c+di)\quad\text{here}\quad i=\sqrt{-1}; a,b,c,d,\alpha\in\mathbb{R}.
$$
Then for $c+di\neq 0$, we have
$$
\frac{a+bi}{c+di}=\alpha(a+bi)(c-di)\quad\text{with}\quad\alpha=\frac{1}{c^2+d^2}.
$$
In fact, this is usually how we define division by a nonzero complex number.

In your case, $a,b,c$ and $d$ are all given so just plug in the numbers. And with $a,b,c$ and $d$ being trig functions, I'm sure some simplication is going to happen.
$$z_{1}=2(cos(\frac{pi}{3})+i sin (\frac{pi}{3}) )=2e^{i\frac{pi}{3}}\\z_{2}=1(cos(\frac{pi}{6})-i sin (\frac{pi}{6}) )=1(cos(\frac{pi}{6})
+i sin (\frac{-pi}{6}) )=\\as-we-know\\cos(a)=cos(-a)\\1(cos(\frac{-pi}{6})-i sin (\frac{-pi}{6}) )=1e^{\frac{-pi}{6}\\
$$