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`z`, `z₁`, `z₂` and `z₃` are for complex numbers on a circle with center on (0, 0) and radius = 1. Suppose the angle between the line between `z` and `z₁` and the line between `z₂` and `z₃` is 90 degrees, prove that ```zz₁ + z₂z₃ = 0```.
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Write z in polar form.
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Oops.  I misread things.  But I can salvage my argument by noting that the line through the origin and the midpoint between z_0 and z_1 (“the line separating z_0 and z_1”) is perpendicular to the line joining z_0 and z_1.  So if the lines in your problem are perpendicular, then so are my lines.
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I’d first prove it in the special case when the two lines are parallel to the real or imaginary axes.  Then show that rotating the circle (multiplying all four quantities by the same value w on the unit circle does this) has the effect of multiplying z_0 z_1 - z_2 z_3 by w^2.  If it starts out zero, it remains zero.
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Here’s a quicker way:  Suppose the line separating z_0 and z_1 is at angle θ.  This means z_0 is e^i(θ+α) and z_1 is e^i(θ-α). This means z_0 z_1 = e^2iθ .  Do the same for the other two values, supposing their separating line is at angle θ + π/2.  You’ll see that z_2 z_3 = e^i(2θ+π).  The result follows quickly.
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from 90° condition => (z-z1)/(z2-z3) is imaginary

(z-z1)/(z2-z3) = −[(z-z1)/(z2-z3)]\* = −(z\*-z1\*)/(z2\*-z3\*)

(z-z1)(z2\*-z3\*) + (z\*-z1\*)(z2-z3) = 0  | for numbers on unit circle z*=1/z

(z-z1)(1/z2-1/z3) + (1/z-1/z1)(z2-z3) = 0   | multiply by z z1 z2 z3

z z1(z-z1)(z3-z2) + z2 z3 (z1-z)(z2-z3) = 0

z z1 + z2 z3 = 0
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