Sec. 10.7 Orientation of a triple of non-collinear points 161
and so
z′ 3 −z′ 2 =(z ̄ 3 −z ̄ 2 )cis 2α,z′ 4 −z′ 2 =(z ̄ 4 −z ̄ 2 )cis 2α.
Hence
z′ 4 −z′ 2
z′ 3 −z′ 2
=
z 4 −z 2
z 3 −z 2,
so that
ℑz′ 4 −z′ 2
z′ 3 −z′ 2=−ℑ
z 4 −z 2
z 3 −z 2,
and so by 10.4.1(ii) the result follows.
LetF,F 1 be frames of reference and Z 3 ,Z 4 ,Z 5 non-collinear points. Letθ=
FZ 4 Z 3 Z 5 andφ=F 1 Z 4 Z 3 Z 5 .Then|φ|◦is equal to|θ|◦or 360 −|θ|◦, according
asF 1 is positively or negatively oriented with respect toF.
Proof. We use the notation of 10.6.1. WhenF 1 is positively oriented with respect
toF, we recall that forf(Z)=Z′withz′=z 0 +zcisα,wehavef(F)=F 1 .On
solving this forzand then interchangingzandz′, we see that
f−^1 (Z)∼F(z−z 0 )cis(−α).Then by 8.2.1(xii),Z=f(f−^1 (Z))∼F 1 (z−z 0 )cis(−α).
LettingZj∼Fzj,Z′j∼F 1 z′jwe then havez′j=(zj−z 0 )cis(−α).Thus
z′ 5 −z′ 3
z′ 4 −z′ 3=
(z 5 −z 0 )cis(−α)−(z 3 −z 0 )cis(−α)
(z 4 −z 0 )cis(−α)−(z 3 −z 0 )cis(−α)=
z 5 −z 3
z 4 −z 3.
But by 10.4.1(i),
z 5 −z 3
z 4 −z 3=
|Z 3 ,Z 5 |
|Z 3 ,Z 4 |
cisθ,z′ 5 −z′ 3
z′ 4 −z′ 3=
|Z 3 ,Z 5 |
|Z 3 ,Z 4 |
cisφ.Thus cisφ=cisθand so|φ|◦=|θ|◦.
WhenF 1 is negatively oriented with respect toF, we take instead f(Z)=Z′
withz′=z 0 +z ̄cisα.Nowf−^1 (Z)∼F( ̄z−z ̄ 0 )cisαand so
z′ 5 −z′ 3
z′ 4 −z′ 3=(z ̄^5 −z ̄^0 )cis(α)−(z ̄^3 −z ̄^0 )cis(α)
(z ̄ 4 −z ̄ 0 )cis(α)−(z ̄ 3 −z ̄ 0 )cis(α)=z^5 −z^3
z 4 −z 3.
Thus cisφ=cisθ=cis(−θ)and so|φ|◦=|(−θ)|◦= 360 −|θ|◦.
LetFandF 1 be frames of reference. Then the ratios of complex-valued distancesρ=Z 3 Z 4 F
Z 1 Z 2 F
, σ=Z 3 Z 4 F 1
Z 1 Z 2 F 1
,
defined in 10.2.1, satisfyσ=ρwhenF 1 is positively oriented with respect toF,
andσ=ρ ̄whenF 1 is negatively oriented with respect toF.