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Linear Algebra Solutions 60

Linear Algebra Solutions 60 - µ z 3-z 1 z 3-z 2 -Arg µ z...

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where t is real. Then 0 < t < 1 describes the segment ab . Also z - a z - b = t t - 1 . Hence z - a z - b is real and negative if z is on the segment a , but is real and positive if z is on the remaining part of the line, with corresponding values Arg z - a z - b = π, 0 , respectively. (ii) Case (a) Suppose z 1 , z 2 and z 3 are not collinear. Then these points determine a circle. Now z 1 and z 2 partition this circle into two arcs. If z 3 and z 4 lie on the same arc, then Arg z 3 - z 1 z 3 - z 2 = Arg z 4 - z 1 z 4 - z 2 ; whereas if z 3 and z 4 lie on opposite arcs, then Arg z 3 - z 1 z 3 - z 2 = α and Arg z 4 - z 1 z 4 - z 2 = α - π. Hence in both cases Arg z 3 - z 1 z 3 - z 2 / z
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Unformatted text preview: µ z 3-z 1 z 3-z 2 ¶-Arg µ z 4-z 1 z 4-z 2 ¶ (mod 2 π ) ≡ 0 or π. In other words, the cross–ratio z 3-z 1 z 3-z 2 / z 4-z 1 z 4-z 2 is real. (b) If z 1 , z 2 and z 3 are collinear, then again the cross–ratio is real. The argument is reversible. (iii) Assume that A, B, C, D are distinct points such that the cross–ratio r = z 3-z 1 z 3-z 2 / z 4-z 1 z 4-z 2 is real. Now r cannot be 0 or 1. Then there are three cases: 63...
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