Epimorphism from Real Numbers to Circle Group

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Theorem

Let $\struct {K, \times}$ be the circle group, that is:

$K = \set {z \in \C: \cmod z = 1}$

under complex multiplication.

Let $f: \R \to K$ be the mapping from the real numbers to $K$ defined as:

$\forall x \in \R: \map f x = \cos x + i \sin x$


Then $f: \struct {\R, +} \to \struct {K, \times}$ is a group epimorphism.


Its kernel is:

$\map \ker f = \set {2 \pi n: n \in \Z}$


Proof

$f$ is a surjection from ...


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$\Box$


Then:

\(\ds \map f x \times \map f y\) \(=\) \(\ds \paren {\cos x + i \sin x} \paren {\cos y + i \sin y}\)
\(\ds \) \(=\) \(\ds \cos x \cos y + i \sin x \cos y + \cos x i \sin y + i \sin x i \sin y\)
\(\ds \) \(=\) \(\ds \paren {\cos x \cos y - \sin x \sin y} + i \paren {\sin x \cos y + \cos x \sin y}\) as $i^2 = -1$
\(\ds \) \(=\) \(\ds \map \cos {x + y} + i \map \sin {x + y}\) Cosine of Sum and Sine of Sum
\(\ds \) \(=\) \(\ds \map f {x + y}\)

So $f$ is a (group) homomorphism.

$\Box$


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

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

$\Box$


From Cosine of Multiple of Pi:

$\forall n \in \Z: \cos n \pi = \paren {-1}^n$

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 $\cos x + i \sin x = 1 + 0 i$ 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:

$x \in \set {2 \pi n: n \in \Z}$

Hence, by definition of kernel:

$\map \ker f = \set {2 \pi n: n \in \Z}$

$\blacksquare$


Sources

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