# Cosets of Positive Reals in Multiplicative Group of Complex Numbers

## Theorem

Let $S$ be the positive real axis of the complex plane:

$S = \set {z \in \C: z = x + 0 i, x \in \R_{>0} }$

Consider the algebraic structure $\struct {S, \times}$ as a subgroup of the multiplicative group of complex numbers $\struct {\C_{\ne 0}, \times}$.

The cosets of $\struct {S, \times}$ are the sets of the form:

$\set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$ for some $\theta \in \hointr 0 {2 \pi}$

That is, the sets of all complex numbers with a constant argument.

## Proof

Let $z_0 \in \C_{\ne 0}$.

Write $z_0 = r_0 e^{i \theta}$, where $r_0 > 0$ and $\theta \in \hointr 0 {2 \pi}$.

We will show that:

$z_0 S = \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$

Pick any $w \in z_0 S$.

Then there exists some $x \in S$ such that $w = z_0 x$.

Note that $x \in \R_{>0}$ and $r_0 x \in \R_{>0}$.

Thus:

$w = z_0 x = r_0 x e^{i \theta} \in \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$

This shows that:

$z_0 S \subseteq \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$

Pick any $w \in \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$.

Write $w = r e^{i \theta}$, where $r \in \R_{>0}$.

Note that $\dfrac r {r_0} \in \R_{>0}$.

Then:

$w = \paren {\dfrac r {r_0}} r_0 e^{i \theta} = \paren {\dfrac r {r_0}} z_0 \in z_0 S$

This shows that:

$z_0 S \supseteq \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$

By definition of set equality:

$z_0 S = \set {z \in \C: \exists r \in \R_{>0}: z = r e^{i \theta}}$

$\blacksquare$