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Linear Algebra Solutions 61

# Linear Algebra Solutions 61 - (i 0< r< 1(ii r< 0(iii...

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Unformatted text preview: (i) 0 < r < 1; (ii) r < 0; (iii) r > 1. Case (i). Here |r| + |1 β r| = 1. So z4 β z 1 z3 β z 2 + 1β Β· z4 β z 2 z3 β z 1 z4 β z 1 z3 β z 2 Β· z4 β z 2 z3 β z 1 = 1. Multiplying both sides by the denominator |z4 β z2 ||z3 β z1 | gives after simpliο¬cation |z4 β z1 ||z3 β z2 | + |z2 β z1 ||z4 β z3 | = |z4 β z2 ||z3 β z1 |, or (a) AD Β· BC + AB Β· CD = BD Β· AC. Case (ii). Here 1 + |r| = |1 β r|. This leads to the equation (b) BD Β· AC + AD Β· BC + = AB Β· CD. Case (iii). Here 1 + |1 β r| = |r|. This leads to the equation (c) BD Β· AC + AB Β· CD = AD Β· 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 β r| = 1, 1 + |r| = |1 β r|, from which we deduce that r is real. 64 1 + |1 β r| = |r|, ...
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