Complex Numbers form Algebra

Theorem
The set of complex numbers $\C$ forms an algebra over the Field of Real Numbers.

This algebra is:
 * $(1): \quad$ An associative algebra.
 * $(2): \quad$ A commutative algebra.
 * $(3): \quad$ A normed division algebra.
 * $(4): \quad$ A nicely normed $*$-algebra whose $*$ operator is the identity mapping.

Construction of Algebra
We have that $\left({\R, +, \times}\right)$ is a field.

Let this be expressed as $\left({\R, +_\R, \times_\R}\right)$ in order to call attention to the precise scope of the operators.

From Real Numbers form Vector Space, we have that $\left({\R^2, +, \cdot}\right)_\R$ is a vector space, where:
 * the field is $\left({\R, +_R, \times_R}\right)$
 * the abelian group is $\left({\R^2, +_G}\right)$ where $+_G$ is real addition on the cartesian product $\R^2$:
 * $\forall \mathbf a = \left({a_1, a_2}\right), \mathbf b = \left({b_1, b_2}\right) \in \R^2: \mathbf a + \mathbf b = \left({a_1 + b_1, a_2 + b_2}\right)$

Let $\mathbf x = \left({x_1, x_2}\right), \mathbf y = \left({y_1, y_2}\right)$.

Let $\times$ be the binary operation on $\left({\R^2, +, \cdot}\right)_\R$ defined as:
 * $\forall \mathbf x, \mathbf y \in \left({\R^2, +, \cdot}\right)_\R: \mathbf x \times \mathbf y = \left({x_1 \times_R y_1 - y_2 \times_R x_2, x_1 \times_R y_2 + y_1 \times_R x_2}\right)$

where $\times_R$ is real multiplication.

Thus $\mathbf x \times \mathbf y$ has been defined as complex multiplication.

It is seen that this structure as presented is the same as that set forth in Complex Numbers: Formal Definition.

Proof of an Algebra
We need to show that $\times$ as defined on $\left({\R^2, +, \cdot}\right)_\R$ as:
 * $\forall x, y \in \left({\R^2, +, \cdot}\right)_\R = x \times_R y$

is bilinear.

That is: $\forall a, b \in \R, x, y \in \R^2$:
 * $\left({\left({a \cdot x}\right) + \left({b \cdot y}\right)}\right) \times z = \left({a \cdot \left({x \times z}\right)}\right) + \left({b \cdot \left({y \times z}\right)}\right)$
 * $z \times \left({\left({a \cdot x}\right) + \left({b \cdot y}\right)}\right) = \left({a \cdot \left({z \times x}\right)}\right) + \left({b \cdot \left({z \times y}\right)}\right)$

So:

Similarly:

So the set of complex numbers forms an algebra $\left({\R, \times}\right)$.

In order to simplify the notation, $\times$ is to be used to denote both real multiplication and the complex multiplication operation which has just been defined in $\left({\R, \times}\right)$.

Which one is meant at any particular point is to be determined by context.

Proof of Associativity
This has been demonstrated in Complex Multiplication is Associative.

Proof of Commutativity
This has been demonstrated in Complex Multiplication is Commutative.

Proof of Normed Division Algebra
Consider the element $\left({1, 0}\right)$ of $\R^2$.

We have:

As $\times$ has already been shown to be commutative, it follows that $\left({1, 0}\right) \times \left({x_1, x_2}\right) = \left({x_1, x_2}\right)$.

So $\left({1, 0}\right) \in \R^2$ functions as a unit.

That is, $\left({\R^2, \times}\right)$ is a unitary algebra.

We define a norm on $\left({\R^2, \times}\right)$ by:
 * $\forall \mathbf a = \left({a_1, a_2}\right) \in \R^2: \left \Vert {\mathbf a} \right \Vert = \sqrt {a_1^2 + a_2^2}$

This is a norm because:


 * $(1): \quad \left \Vert \mathbf x \right \Vert = 0 \iff \mathbf x = \mathbf 0$
 * $(2): \quad \left \Vert \lambda \mathbf x \right \Vert = \left \vert \lambda \right \vert \left \Vert x \right \Vert$
 * $(3): \quad \left \Vert x - y \right \Vert \le \left \Vert x - z \right \Vert + \left \Vert z - y \right \Vert$

It also follows that:
 * $\left \Vert x \times y \right \Vert = \left \vert x \times y \right \vert = \left \vert x \right \vert \times \left \vert y \right \vert = \left \Vert x \right \Vert \times \left \Vert y \right \Vert$

and so $\left({\R^2, \times}\right)$ is a normed division algebra.

Proof of Nicely Normed $*$-Algebra
We define the conjugation $*$ as follows:


 * $\forall \mathbf a = \left({a_1, a_2}\right) \in \R^2: \mathbf a^* = \left({a_1, -a_2}\right)$

We have that:

demonstrating that $*$ is indeed a conjugation.

Then we have that:

Similarly for $\mathbf a^* + \mathbf a$.

So both $a^* + a$ and $a \times a^*$ are real.

So $\left({\R^2, \times}\right)$ is a nicely normed $*$-algebra.