# Convergence of Complex Sequence in Polar Form

## Theorem

Let $z \ne 0$ be a complex number with modulus $r$ and argument $\theta$.

Let $\left\langle{z_n}\right\rangle$ be a sequence of nonzero complex numbers.

Let $r_n$ be the modulus of $z_n$ and $\theta_n$ be an argument of $z_n$.

Then $z_n$ converges to $z$ if and only if the following hold:

$(1): \quad r_n$ converges to $r$
$(2): \quad$ There exists a sequence $(k_n)$ of integers such that $\theta_n + 2 k_n \pi$ converges to $\theta$.

### Corollary

Let $\left\langle{z_n}\right\rangle$ be a sequence of nonzero complex numbers.

Let $z \ne 0$ 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 $I$.

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 $z$ if and only if $r_n$ converges to $r$ and $\theta_n$ converges to $\theta$.

## Proof

Suppose $r_n \to r$ and $\theta_n + 2 k_n \pi \to \theta$.

We have, by Complex Modulus of Difference of Complex Numbers:

 $\displaystyle \left\vert{z_n - z}\right\vert^2$ $=$ $\displaystyle r_n^2 + r^2 - 2 r r_n \cos \left({\theta_n + 2 k_n \pi - \theta}\right)$

Because Cosine Function is Continuous, $\cos \left({\theta_n + 2 k_n \pi - \theta}\right) \to 1$.

It follows that $\left\vert{z_n - z}\right\vert^2 \to 0$.

Conversely, suppose $z_n \to z$.

By Modulus of Limit, $r_n \to r$.

We have, by Complex Modulus of Difference of Complex Numbers:

 $\displaystyle \cos \left({\theta_n - \theta}\right)$ $=$ $\displaystyle \frac {r_n^2 + r^2 - \left\vert{z_n - z}\right\vert^2} {2 r r_n} \to 1$

By Convergence of Cosine of Sequence, there exists a sequence $\left\langle{k_n}\right\rangle$ of integers such that $\theta_n + 2 k_n \pi$ converges to $\theta$.

$\blacksquare$