Circle Group is Group/Proof 2

Proof
We note that $K \ne \O$ as the identity element $1 + 0 i \in K$.

Since all $z \in K$ have modulus $1$, they have, for some $\theta \in \hointr 0 {2 \pi}$, the polar form:


 * $z = \map \exp {i \theta} = \cos \theta + i \sin \theta$

Conversely, if a complex number has such a polar form, it has modulus $1$.

Observe the following property of the complex exponential function:


 * $\forall a, b \in \C: \map \exp {a + b} = \map \exp a \map \exp b$

We must show that if $x, y \in K$ then $x \cdot y^{-1} \in K$.

Let $x, y \in K$ be arbitrary.

Choose suitable $s, t \in \hointr 0 {2 \pi}$ such that:


 * $x = \map \exp {i s}$
 * $y = \map \exp {i t}$

We compute:

So $y^{-1} = \map \exp {-i t}$.

We note that this lies in $K$.

Furthermore, we have:

We conclude that $x y \in K$.

By the Two-Step Subgroup Test, $K$ is a subgroup of $\C$ under complex multiplication.