Definition 82 (Landau 66). For a positive Ξ, √Ξ is the positive solution of ξ • ξ = Ξ, which exists and is one only, by Theorem 255.
Absolute Value
Definitions 82–84 · Theorems 396–405
Definition 83 (Landau 67). √0 = 0.
Definition 84 (Landau 68). |[Ξ₁, Ξ₂]| = √(Ξ₁ • Ξ₁ + Ξ₂ • Ξ₂). It is the absolute value of the complex number.
It is defined, since Ξ₁ • Ξ₁ + Ξ₂ • Ξ₂ ≧ 0 by Definition 71.
Theorem 396 (Landau 264). |x| > 0 for x ≠ o, and |x| = 0 for x = o.
Definitions 82 to 84.
Lemma. If Ξ ≧ 0, Θ ≧ 0 and Ξ • Ξ ≧ Θ • Θ, then Ξ ≧ Θ.
Otherwise Ξ < Θ, so Θ > 0, and Ξ • Ξ ≦ Ξ • Θ < Θ • Θ by Theorem 337.
Theorem 397 (Landau 265). |[Ξ₁, Ξ₂]| ≧ |Ξ₁| and |[Ξ₁, Ξ₂]| ≧ |Ξ₂|.
|x| • |x| = Ξ₁ • Ξ₁ + Ξ₂ • Ξ₂, which is ≧ Ξ₁ • Ξ₁ = |Ξ₁| • |Ξ₁| and ≧ Ξ₂ • Ξ₂ = |Ξ₂| • |Ξ₂|. Then the Lemma.
Theorem 398 (Landau 266). If [Ξ, 0] • [Ξ, 0] = [Θ, 0] • [Θ, 0], with Ξ ≧ 0 and Θ ≧ 0, then Ξ = Θ.
[Λ, 0] • [Λ, 0] = [Λ • Λ, 0]. So Ξ • Ξ = Θ • Θ, and by the Lemma both Ξ ≧ Θ and Θ ≧ Ξ.
Theorem 399 (Landau 267). [|x|, 0] • [|x|, 0] = x • x*.
[|x| • |x|, 0] = [Ξ₁Ξ₁ + Ξ₂Ξ₂, 0] = [Ξ₁Ξ₁ − Ξ₂(−Ξ₂), Ξ₁(−Ξ₂) + Ξ₂Ξ₁] = x • x*.
Theorem 400 (Landau 268). |x • y| = |x| • |y|.
Proof: By Theorems 399 and 393,
[|x • y|, 0] • [|x • y|, 0] = (x • y) • (x* • y*) = (x • x*) • (y • y*) = [|x| • |x|, 0] • [|y| • |y|, 0] = [|x| • |y|, 0] • [|x| • |y|, 0].
Then Theorem 398.
Theorem 401 (Landau 269). If y ≠ o, then |x/y| = |x|/|y|.
|y| > 0. From (x/y) • y = x and Theorem 400, |x/y| • |y| = |x|.
Theorem 402 (Landau 270). If x + y = e, then |x| + |y| ≧ 1.
By Theorem 397, |x| ≧ |Ξ₁| ≧ Ξ₁ and |y| ≧ |Θ₁| ≧ Θ₁. So |x| + |y| ≧ Ξ₁ + Θ₁ = 1.
Theorem 403 (Landau 271). |x + y| ≦ |x| + |y|.
Proof: If x + y = o, the left side is 0. Otherwise x/(x + y) + y/(x + y) = e by Theorems 385 and 382, so by Theorems 402 and 401, |x|/|x + y| + |y|/|x + y| ≧ 1. Multiply by |x + y| > 0 (Theorem 337).
Theorem 404 (Landau 272). |−x| = |x|.
(−Ξ₁)(−Ξ₁) + (−Ξ₂)(−Ξ₂) = Ξ₁Ξ₁ + Ξ₂Ξ₂.
Theorem 405 (Landau 273). |x − y| ≧ | |x| − |y| |.
Proof: x = y + (x − y), so |x| ≦ |y| + |x − y| by Theorem 403, and |x − y| ≧ |x| − |y|. Exchanging x and y, and using Theorem 404, |x − y| = |y − x| ≧ |y| − |x| = −(|x| − |y|). The absolute value of |x| − |y| is one of these two.
Note that complex numbers are not ordered. Greater and less are said only of their absolute values, which are real.