Complex Numbers Form a Field

A formal proof that the set of complex numbers satisfies all field axioms

Introduction

The set of complex numbers, denoted by ℂ, consists of all numbers of the form \(a + bi\), where \(a, b \in \mathbb{R}\) and \(i\) is the imaginary unit with \(i^2 = -1\).

We will prove that ℂ satisfies all the axioms of a field, making it one of the most important algebraic structures in mathematics.

Complex Number Representation

A complex number \(z\) is written as \(z = a + bi\), where:

- \(a\) is the real part (Re(z))

- \(b\) is the imaginary part (Im(z))

- \(i\) is the imaginary unit with the property \(i^2 = -1\)

Field Axioms and Their Verification

Closure under Addition

Axiom 1

For any two complex numbers \(z = a + bi\) and \(w = c + di\):

\(z + w = (a + c) + (b + d)i\)

Since \(a, b, c, d \in \mathbb{R}\), \(a + c\) and \(b + d\) are also real numbers. Thus, \(z + w \in \mathbb{C}\).

Closure under Multiplication

Axiom 2

For \(z = a + bi\) and \(w = c + di\):

\(z \cdot w = (a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i\)

Since \(ac - bd\) and \(ad + bc\) are real numbers, \(z \cdot w \in \mathbb{C}\).

Associativity of Addition

Axiom 3

For \(z = a + bi\), \(w = c + di\), and \(v = e + fi\):

\((z + w) + v = [(a + c) + (b + d)i] + (e + fi) = (a + c + e) + (b + d + f)i\)

\(z + (w + v) = (a + bi) + [(c + e) + (d + f)i] = (a + c + e) + (b + d + f)i\)

Thus, \((z + w) + v = z + (w + v)\).

Associativity of Multiplication

Axiom 4

Using the properties of real numbers:

\((z \cdot w) \cdot v = z \cdot (w \cdot v)\)

This can be verified by expanding both sides, but it holds due to the associativity of real number operations.

Commutativity of Addition

Axiom 5

\(z + w = (a + c) + (b + d)i = (c + a) + (d + b)i = w + z\)

Commutativity of Multiplication

Axiom 6

\(z \cdot w = (ac - bd) + (ad + bc)i = (ca - db) + (cb + da)i = w \cdot z\)

Existence of Additive Identity

Axiom 7

The complex number \(0 = 0 + 0i\) satisfies:

\(z + 0 = (a + bi) + (0 + 0i) = a + bi = z\)

Thus, \(0\) is the additive identity.

Existence of Multiplicative Identity

Axiom 8

The complex number \(1 = 1 + 0i\) satisfies:

\(z \cdot 1 = (a + bi)(1 + 0i) = a \cdot 1 + a \cdot 0i + bi \cdot 1 + bi \cdot 0i = a + bi = z\)

Thus, \(1\) is the multiplicative identity.

Existence of Additive Inverses

Axiom 9

For \(z = a + bi\), the additive inverse is \(-z = -a - bi\), since:

\(z + (-z) = (a + bi) + (-a - bi) = 0 + 0i = 0\)

Existence of Multiplicative Inverses

Axiom 10

For any non-zero \(z = a + bi\) (so \(a^2 + b^2 \neq 0\)), the multiplicative inverse is:

\(z^{-1} = \frac{a - bi}{a^2 + b^2}\)

because:

\(z \cdot z^{-1} = (a + bi) \cdot \frac{a - bi}{a^2 + b^2} = \frac{a^2 + b^2}{a^2 + b^2} = 1\)

Distributivity of Multiplication over Addition

Axiom 11

For \(z = a + bi\), \(w = c + di\), and \(v = e + fi\):

\(z \cdot (w + v) = (a + bi)[(c + e) + (d + f)i] = a(c + e) - b(d + f) + [a(d + f) + b(c + e)]i\)

\(= (ac + ae - bd - bf) + (ad + af + bc + be)i\)

Similarly,

\(z \cdot w + z \cdot v = [(ac - bd) + (ad + bc)i] + [(ae - bf) + (af + be)i] = (ac - bd + ae - bf) + (ad + bc + af + be)i\)

Thus, \(z \cdot (w + v) = z \cdot w + z \cdot v\).

Conclusion

Since all field axioms are satisfied, we conclude that the complex numbers ℂ form a field.

This proof demonstrates that complex numbers are a well-defined algebraic structure, which is fundamental for more advanced topics in mathematics, such as complex analysis or polynomial theory.

Additional Note

This proof demonstrates that complex numbers are a well-defined algebraic structure, which is fundamental for more advanced topics in mathematics, such as complex analysis or polynomial theory.

Complex Numbers Field Proof | Mathematical Foundations

Created as an educational resource for understanding algebraic structures