Convergence of Complex Sequence in Polar Form

Theorem

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

Let $\sequence {z_n}$ 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 $\sequence {k_n}$ of integers such that $\theta_n + 2 k_n \pi$ converges to $\theta$.

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 $\theta$.

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

$\ds \cmod {z_n - z}^2$

$=$

$\ds r_n^2 + r^2 - 2 r r_n \, \map \cos {\theta_n + 2 k_n \pi - \theta}$

$\map \cos {\theta_n + 2 k_n \pi - \theta} \to 1$

It follows that:

$\cmod {z_n - z}^2 \to 0$

Conversely, suppose $z_n \to z$.

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

$\ds \map \cos {\theta_n - \theta}$

$=$

$\ds \frac {r_n^2 + r^2 - \cmod {z_n - z}^2} {2 r r_n} \to 1$

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

$\blacksquare$