Convergence of Complex Sequence in Polar Form/Corollary

Corollary to Convergence of Complex Sequence in Polar Form
Let $z\neq0$ be a complex number with modulus $r$ and argument $\theta$.

Let $I$ be a real interval of length at most $2\pi$ that contains $\theta$.

Suppose $\theta$ is not an endpoint of $I$.

Let $(z_n)$ be a sequence of nonzero complex numbers.

Suppose each $z_n$ admits an argument $\theta_n\in I$.

Let $r_n$ be the modulus of $z_n$.

Then $z_n$ converges to $z$ iff $r_n$ converges to $r$ and $\theta_n$ converges to $\theta$.

Proof
Suppose $r_n\to r$ and $\theta_n\to\theta$.

Then by Convergence of Complex Sequence in Polar Form, $z_n\to z$.

Conversely, suppose $z_n\to z$.

By Convergence of Complex Sequence in Polar Form, we have:
 * $r_n\to r$
 * There exists a sequence $(k_n)$ of integers such that $\theta_n+2k_n\pi$ converges to $\theta$.

It remains to prove that $\theta_n\to\theta$.

Let $N\in\N$ such that $|\theta_N+2k_N\pi-\theta|\leq\pi/2$ for all $n\geq N$.

By the Triangle Inequality:
 * $|2k_n\pi-2k_N\pi|\leq|\theta_n+2k_n\pi-\theta|+|\theta_N+2k_N\pi-\theta|\leq\pi$

for all $n\geq N$.

Thus $|k_n-k_N|\leq1/2$, so $k_n=k_N$ for all $n\geq N$.

By the Triangle Inequality:
 * $|2\pi k_N|\leq|\theta_n-\theta|+|\theta_n+2\pi k_N-\theta|$ for all $n\in\N$.

Because $\theta_n\in I$ and $\theta$ is not an endpoint of $I$, $|\theta_n-\theta|<2\pi$ for all $n\in\N$.

Because $\theta_n+2\pi k_N-\theta\to0$, $|2\pi k_N|<2\pi$.

Thus $k_n=0$ for all $n\geq N$.

Thus $\theta_n\to\theta$.