Circle Group is Group/Proof 1

Proof
First we note that $K \subseteq \C$.

So to show that $K$ is a group it is sufficient to show that $K$ is a subgroup of the multiplicative group of complex numbers $\struct {\C_{\ne 0}, \times}$.

From Complex Multiplication Identity is One, the identity element $1 + 0 i$ is in $K$.

Thus $K \ne \O$.

We now show that $z, w \in K \implies z w \in K$:

Next we see that $z \in K \implies z^{-1} \in K$:


 * $\cmod z = 1 \implies \cmod {\dfrac 1 z} = 1$

Thus by the Two-Step Subgroup Test:
 * $K \le \C_{\ne 0}$

Thus $K$ is a group.