Linear Algebra Solutions 61

Linear Algebra Solutions 61 - (i) 0 < r < 1;...

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Unformatted text preview: (i) 0 < r < 1; (ii) r < 0; (iii) r > 1. Case (i). Here |r| + |1 − r| = 1. So z4 − z 1 z3 − z 2 + 1− · z4 − z 2 z3 − z 1 z4 − z 1 z3 − z 2 · z4 − z 2 z3 − z 1 = 1. Multiplying both sides by the denominator |z4 − z2 ||z3 − z1 | gives after simplification |z4 − z1 ||z3 − z2 | + |z2 − z1 ||z4 − z3 | = |z4 − z2 ||z3 − z1 |, or (a) AD · BC + AB · CD = BD · AC. Case (ii). Here 1 + |r| = |1 − r|. This leads to the equation (b) BD · AC + AD · BC + = AB · CD. Case (iii). Here 1 + |1 − r| = |r|. This leads to the equation (c) BD · AC + AB · CD = AD · BC. Conversely if (a), (b) or (c) hold, then we can reverse the argument to deduce that r is a complex number satisfying one of the equations |r| + |1 − r| = 1, 1 + |r| = |1 − r|, from which we deduce that r is real. 64 1 + |1 − r| = |r|, ...
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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