Metamath Proof Explorer


Theorem absreimsq

Description: Square of the absolute value of a number that has been decomposed into real and imaginary parts. (Contributed by NM, 1-Feb-2007)

Ref Expression
Assertion absreimsq ⊢ A ∈ ℝ ∧ B ∈ ℝ → A + i ⁢ B 2 = A 2 + B 2

Proof

Step Hyp Ref Expression
1 recn ⊢ A ∈ ℝ → A ∈ ℂ
2 ax-icn ⊢ i ∈ ℂ
3 recn ⊢ B ∈ ℝ → B ∈ ℂ
4 mulcl ⊢ i ∈ ℂ ∧ B ∈ ℂ → i ⁢ B ∈ ℂ
5 2 3 4 sylancr ⊢ B ∈ ℝ → i ⁢ B ∈ ℂ
6 addcl ⊢ A ∈ ℂ ∧ i ⁢ B ∈ ℂ → A + i ⁢ B ∈ ℂ
7 1 5 6 syl2an ⊢ A ∈ ℝ ∧ B ∈ ℝ → A + i ⁢ B ∈ ℂ
8 absvalsq2 ⊢ A + i ⁢ B ∈ ℂ → A + i ⁢ B 2 = ℜ ⁡ A + i ⁢ B 2 + ℑ ⁡ A + i ⁢ B 2
9 7 8 syl ⊢ A ∈ ℝ ∧ B ∈ ℝ → A + i ⁢ B 2 = ℜ ⁡ A + i ⁢ B 2 + ℑ ⁡ A + i ⁢ B 2
10 crre ⊢ A ∈ ℝ ∧ B ∈ ℝ → ℜ ⁡ A + i ⁢ B = A
11 10 oveq1d ⊢ A ∈ ℝ ∧ B ∈ ℝ → ℜ ⁡ A + i ⁢ B 2 = A 2
12 crim ⊢ A ∈ ℝ ∧ B ∈ ℝ → ℑ ⁡ A + i ⁢ B = B
13 12 oveq1d ⊢ A ∈ ℝ ∧ B ∈ ℝ → ℑ ⁡ A + i ⁢ B 2 = B 2
14 11 13 oveq12d ⊢ A ∈ ℝ ∧ B ∈ ℝ → ℜ ⁡ A + i ⁢ B 2 + ℑ ⁡ A + i ⁢ B 2 = A 2 + B 2
15 9 14 eqtrd ⊢ A ∈ ℝ ∧ B ∈ ℝ → A + i ⁢ B 2 = A 2 + B 2