Metamath Proof Explorer


Theorem absval

Description: The absolute value (modulus) of a complex number. Proposition 10-3.7(a) of Gleason p. 133. (Contributed by NM, 27-Jul-1999) (Revised by Mario Carneiro, 7-Nov-2013)

Ref Expression
Assertion absval ( 𝐴 ∈ ℂ → ( abs ‘ 𝐴 ) = ( √ ‘ ( 𝐴 · ( ∗ ‘ 𝐴 ) ) ) )

Proof

Step Hyp Ref Expression
1 fveq2 ( 𝑥 = 𝐴 → ( ∗ ‘ 𝑥 ) = ( ∗ ‘ 𝐴 ) )
2 oveq12 ( ( 𝑥 = 𝐴 ∧ ( ∗ ‘ 𝑥 ) = ( ∗ ‘ 𝐴 ) ) → ( 𝑥 · ( ∗ ‘ 𝑥 ) ) = ( 𝐴 · ( ∗ ‘ 𝐴 ) ) )
3 1 2 mpdan ( 𝑥 = 𝐴 → ( 𝑥 · ( ∗ ‘ 𝑥 ) ) = ( 𝐴 · ( ∗ ‘ 𝐴 ) ) )
4 3 fveq2d ( 𝑥 = 𝐴 → ( √ ‘ ( 𝑥 · ( ∗ ‘ 𝑥 ) ) ) = ( √ ‘ ( 𝐴 · ( ∗ ‘ 𝐴 ) ) ) )
5 df-abs abs = ( 𝑥 ∈ ℂ ↦ ( √ ‘ ( 𝑥 · ( ∗ ‘ 𝑥 ) ) ) )
6 fvex ( √ ‘ ( 𝐴 · ( ∗ ‘ 𝐴 ) ) ) ∈ V
7 4 5 6 fvmpt ( 𝐴 ∈ ℂ → ( abs ‘ 𝐴 ) = ( √ ‘ ( 𝐴 · ( ∗ ‘ 𝐴 ) ) ) )