Epimorphism from Real Numbers to Circle Group

Theorem
Let $\left({K, \times}\right)$ be the circle group, that is:
 * $K = \left\{{z \in \C: \left|{z}\right| = 1}\right\}$

under complex multiplication.

Let $f: \R \to K$ be the mapping from the real numbers to $K$ defined as:
 * $\forall x \in \R: f \left({x}\right) = \cos x + i \sin x$

Then $f: \left({\R, +}\right) \to \left({K, \times}\right)$ is a group epimorphism.

Its kernel is:
 * $\ker \left({f}\right) = \left\{{2 \pi n: n \in \Z}\right\}$

Proof
$f$ is a surjection from ...

Then:

So $f$ is a (group) homomorphism.

Thus $f$ is seen to be a surjective homomorphism.

Hence, by definition, it is a (group) epimorphism.

From Cosine of Multiple of Pi:
 * $\forall n \in \Z: \cos n \pi = \left({-1}\right)^n$

and from Sine of Multiple of Pi:
 * $\forall n \in \Z: \sin n \pi = 0$

From Sine and Cosine are Periodic on Reals, it follows that these are the only values of $\Z$ for which this holds.

For $\cos x + i \sin x = 1 + 0i$ it is necessary that:
 * $\cos x = 1$
 * $\sin x = 0$

and it can be seen that the only values of $x$ for this to happen is:
 * $x \in \left\{{2 \pi n: n \in \Z}\right\}$

Hence, by definition of kernel:
 * $\ker \left({f}\right) = \left\{{2 \pi n: n \in \Z}\right\}$