Maths-
General
Easy

Question

If open vertical bar fraction numerator z subscript 1 end subscript over denominator z subscript 2 end subscript end fraction close vertical bar =1 and arg (z1 z2) = 0, then

  1. z1 = z2    
  2. |z2|2 = z1z2    
  3. z1z2 = 1    
  4. none of these.    

The correct answer is: |z2|2 = z1z2


    Let z1 = r1( cosq1 + i sinq1) then open vertical bar fraction numerator z subscript 1 end subscript over denominator z subscript 2 end subscript end fraction close vertical bar equals 1
    Þ |z1| = |z2| Þ|z1| = |z2| = r1 .
    Now arg (z1 z2) = 0 Þ arg( z1) + arg(z2) = 0 Þ arg(z2) = – q1
    Therefore, z2 = r1 ( cos(–q1) + i sin(–q1)) = r1( cosq1 – i sinq1) = stack z with ̄ on top subscript 1 end subscript
    Þ stack z with ̄ on top subscript 2 end subscript= open parentheses stack stack z with ̄ on top with ̄ on top subscript 1 end subscript close parentheses = z1 Þ|z2|2 = z1 z2 .
    Hence (b) is the correct answer.

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