PROOF 89

Prove.

The product of a complex number and its conjugate is real

On the complex numbers ℂ.

A way to read this proof

First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.

What you may use

A1a(b + c) = ab + ac
distributive law
A2(ab)c = a(bc)
associativity of multiplication
A3ab = ba
commutativity of multiplication
A4a × 1 = a
multiplicative identity
A5(a + b) + c = a + (b + c)
associativity of addition
A6a + b = b + a
commutativity of addition
A7a + 0 = a
additive identity
A8a + (−a) = 0
additive inverse
A9a × 0 = 0
lemma of Proof 01
A10a(−b) = −(ab)
lemma of Proof 14
A11(−a)(−b) = ab
product of negatives (Proof 01)
A12i × i = −1
definition of i (Proof 14)
A13a ≠ 0 ⟹ there exists b with ab = 1
multiplicative inverse of reals
A140 ≤ a × a
the square of a real is nonnegative
A15a ≠ 0 ⟹ 0 < a × a
lemma of Proof 10
A160<a, 0≤b ⟹ 0 < a + b
order and addition
A17exactly one of a<b, a=b, b<a
trichotomy of order
A18p, q ∈ ℝ ⟹ p+q, pq ∈ ℝ
closure of the reals
Ea = b ⟹ apply the same operation to both sides
properties of equality

D1z̄ = a + (−b)i
definition of the complex conjugate
D2|z|² = aa + bb
definition of the absolute value
L1z ≠ 0 ⟹ 0 < aa+bb
Lemma 1
Tz z̄ = aa + bb
Theorem

Nothing outside this list may be cited.

Definition D1

For z = a + bi with a, b real, z̄ = a + (−b)i is called the complex conjugate of z.

Proof.

a − bi means a + (−b)i. The real part is kept; only the sign of the imaginary part changes.convention

∎

Definition D2

|z| is the nonnegative real x satisfying x × x = a × a + b × b.

Proof.

|z| is called the absolute value of z.convention

∎

Lemma L1

If a and b are not both 0, then 0 < a × a + b × b, and this value is not 0.

Proof.

(1)Suppose a ≠ 0.assumption

(2)0 < a × a(1), A15

(3)0 ≤ b × bA14

(4)Therefore 0 < a × a + b × b.(2), (3), A16

(5)In the case b ≠ 0, 0 < b × b and 0 ≤ a × a.A15, A14

(6)a × a + b × b = b × b + a × a, so the same conclusion follows.(5), A6, A16

(7)In either case the value is greater than 0, hence not 0.(4), (6), A17

∎

Theorem T

If z = a + bi, then z z̄ = a × a + b × b, and this value is real.

Proof.

(1)z z̄ = (a + bi)(a + (−b)i)D1

(2)　 = a(a + (−b)i) + (bi)(a + (−b)i)A1

(3)　 = a×a + a((−b)i) + (bi)a + (bi)((−b)i)A1

(4)a((−b)i) = (a(−b))i = (−(ab))iA2, A10

(5)(bi)a = a(bi) = (ab)iA3, A2

(6)(−(ab))i + (ab)i = ((−(ab)) + (ab))i = 0 × i = 0A1, A3, A8, A9

(7)(bi)((−b)i) = (b(−b))(i × i) = (−(b×b))(−1)A2, A3, A10, A12

(8)(−(b×b))(−1) = (b × b) × 1 = b × b(7), A11, A4

(9)Therefore z z̄ = a × a + b × b.(3), (4), (5), (6), (7), (8), A5, A6, A7

(10)a and b are real, so a × a + b × b is real.A18

(11)Therefore z z̄ is real.(9), (10), E

∎

Corollary C

If z ≠ 0, then |z| × |z| = z z̄, and the inverse of z is z̄ divided by a × a + b × b.

Proof.

(1)|z| × |z| = a × a + b × bD2

(2)Therefore |z| × |z| = z z̄.(1), T, E

(3)z ≠ 0, so a and b are not both 0, and a × a + b × b ≠ 0.assumption, D1, L1

(4)There is a real t with (a × a + b × b)t = 1.(3), A13

(5)z(z̄ t) = (z z̄)t = (a × a + b × b)t = 1A2, T, (4), E

(6)Therefore the inverse of z is z̄ t, that is, z̄ divided by a × a + b × b.(5)

∎

Remark

Can A12 be removed from this proof?

Of the four terms, separate the two that cancel from the one that remains. What does the remaining term rest on?