Component wise convergence 
Attempt:
Comparing real and imaginary parts we get:
$r_n \cos \theta_n \to r \cos \theta$
$r_n \sin \theta_n \to r \sin \theta$ $\,\,$ Thus, we get:
$r_n \cos \theta_n \, . \,r_n \cos \theta_n \to r \cos \theta \, . \, r \cos \theta$
$r_n \sin \theta_n \, . \,r_n \sin \theta_n \to r \sin \theta \, . \, r \sin \theta$
$\implies r_n^2 \sin^2 \theta_n \, + \,r_n^2 \cos^2 \theta_n \to r^2 \sin^2 \theta \,+ \, r^2\cos^2 \theta$
$\implies r_n^2  \to r^2$
How do I conclude from here that $r_n \to r$? Should I approach this differently?
Edit: I just realized I could just compose with square root function.
$\lim_{n \to \infty} r_n=\lim_{n \to \infty} \sqrt{{r_n}^2}= \sqrt{\lim_{n \to \infty} {r_n}^2} = \sqrt{r^2}=r$
 A: Sure, given $r_n^2 \to r^2$ and $r_n, r \geq 0$ you can conclude immediately that $r_n \to r$; this follows because the square root function is continuous and thus $r_n = \sqrt{r_n^2} \to \sqrt{r^2} = r$ automatically.
Given this it then follows that $\cos \theta_n \to \cos \theta$ and $\sin \theta_n \to \sin \theta$, so we get $\theta_n \to \theta$, as desired.
Edit from comments: Let me spell out this last part.
First, how do we obtain  $\cos \theta_n \to \cos \theta$ (and likewise for $\sin$)? The argument goes that since $r \not = 0$ and $r_n \to r$ there must exist some $N > 0$ such that for all $n > N$ we have that $r_n \not = 0$. This then allows us to perform the division by $r_n$ (for $n$ sufficiently large) and obtain $\frac{r_n \cos \theta_n}{r_n} \to \frac{r \cos \theta}{r}$.
Once we have $\cos \theta_n \to \cos \theta$, we can compose with the continuous inverse function $\cos^{-1}$ to obtain $\cos^{-1} \cos \theta_n \to \cos^{-1} \cos \theta$. The problem now is just that we do not know the sign of e.g. $\cos^{-1} \cos \theta_n$ (it could be $\theta_n$ or $-\theta_n$). But we can at least conclude that
$$
\lvert \theta_n \rvert = \lvert \cos^{-1} \cos \theta_n \rvert \to \lvert \cos^{-1} \cos \theta \rvert = \lvert \theta \rvert.
$$
If $\theta = 0$ then we are done, since $\lvert \theta_n \rvert \to 0$ if and only if $\theta_n \to 0$. Otherwise we can assume $\theta \not = 0$, and need to argue that the sign of $\theta_n$ is eventually always the same as the sign of $\theta$ (i.e. for $n$ large enough). We do this just using the fact that $\sin \theta_n \to \sin \theta$; we can assume that $\theta_n \not = 0$ as well (by the same argument I gave above for $r_n$), in which case
$$
\operatorname{sign}(\theta_n) = \operatorname{sign}(\sin \theta_n) \to \operatorname{sign}(\sin \theta) = \operatorname{sign}(\theta).
$$
But now we can just multiply this pair of converging sequences and get:
$$
\theta_n = \operatorname{sign}(\theta_n) \lvert \theta_n \rvert \to \operatorname{sign}(\theta) \lvert \theta \rvert = \theta.
$$ (Here $\operatorname{sign}$ is the so-called "sign-function", which is $1$ on positive reals and $-1$ on negative reals.)
