Metamath Proof Explorer


Theorem angrteqvd

Description: Two vectors are at a right angle iff their quotient is purely imaginary. (Contributed by David Moews, 28-Feb-2017)

Ref Expression
Hypotheses ang.1 𝐹 = ( 𝑥 ∈ ( ℂ ∖ { 0 } ) , 𝑦 ∈ ( ℂ ∖ { 0 } ) ↦ ( ℑ ‘ ( log ‘ ( 𝑦 / 𝑥 ) ) ) )
angrteqvd.1 ( 𝜑𝑋 ∈ ℂ )
angrteqvd.2 ( 𝜑𝑋 ≠ 0 )
angrteqvd.3 ( 𝜑𝑌 ∈ ℂ )
angrteqvd.4 ( 𝜑𝑌 ≠ 0 )
Assertion angrteqvd ( 𝜑 → ( ( 𝑋 𝐹 𝑌 ) ∈ { ( π / 2 ) , - ( π / 2 ) } ↔ ( ℜ ‘ ( 𝑌 / 𝑋 ) ) = 0 ) )

Proof

Step Hyp Ref Expression
1 ang.1 𝐹 = ( 𝑥 ∈ ( ℂ ∖ { 0 } ) , 𝑦 ∈ ( ℂ ∖ { 0 } ) ↦ ( ℑ ‘ ( log ‘ ( 𝑦 / 𝑥 ) ) ) )
2 angrteqvd.1 ( 𝜑𝑋 ∈ ℂ )
3 angrteqvd.2 ( 𝜑𝑋 ≠ 0 )
4 angrteqvd.3 ( 𝜑𝑌 ∈ ℂ )
5 angrteqvd.4 ( 𝜑𝑌 ≠ 0 )
6 1 2 3 4 5 angvald ( 𝜑 → ( 𝑋 𝐹 𝑌 ) = ( ℑ ‘ ( log ‘ ( 𝑌 / 𝑋 ) ) ) )
7 6 eleq1d ( 𝜑 → ( ( 𝑋 𝐹 𝑌 ) ∈ { ( π / 2 ) , - ( π / 2 ) } ↔ ( ℑ ‘ ( log ‘ ( 𝑌 / 𝑋 ) ) ) ∈ { ( π / 2 ) , - ( π / 2 ) } ) )
8 4 2 3 divcld ( 𝜑 → ( 𝑌 / 𝑋 ) ∈ ℂ )
9 4 2 5 3 divne0d ( 𝜑 → ( 𝑌 / 𝑋 ) ≠ 0 )
10 8 9 logimclad ( 𝜑 → ( ℑ ‘ ( log ‘ ( 𝑌 / 𝑋 ) ) ) ∈ ( - π (,] π ) )
11 coseq0negpitopi ( ( ℑ ‘ ( log ‘ ( 𝑌 / 𝑋 ) ) ) ∈ ( - π (,] π ) → ( ( cos ‘ ( ℑ ‘ ( log ‘ ( 𝑌 / 𝑋 ) ) ) ) = 0 ↔ ( ℑ ‘ ( log ‘ ( 𝑌 / 𝑋 ) ) ) ∈ { ( π / 2 ) , - ( π / 2 ) } ) )
12 10 11 syl ( 𝜑 → ( ( cos ‘ ( ℑ ‘ ( log ‘ ( 𝑌 / 𝑋 ) ) ) ) = 0 ↔ ( ℑ ‘ ( log ‘ ( 𝑌 / 𝑋 ) ) ) ∈ { ( π / 2 ) , - ( π / 2 ) } ) )
13 8 9 cosarg0d ( 𝜑 → ( ( cos ‘ ( ℑ ‘ ( log ‘ ( 𝑌 / 𝑋 ) ) ) ) = 0 ↔ ( ℜ ‘ ( 𝑌 / 𝑋 ) ) = 0 ) )
14 7 12 13 3bitr2d ( 𝜑 → ( ( 𝑋 𝐹 𝑌 ) ∈ { ( π / 2 ) , - ( π / 2 ) } ↔ ( ℜ ‘ ( 𝑌 / 𝑋 ) ) = 0 ) )