Relations And Functions

  • Question
    CBSEENMA11013052

    Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.

    Solution

    Let     Z1 = (x – iy) (3 + 5i) = 3c + 5xi - 3yi - 5yi2 = (3x + 5y) + i ( 5x - 3y)
       
    Let     Z2    =  -6 -24i
    5
    It is given  that Z1  is conjugate of Z2
     
    rightwards double arrow         Z1     = space space space space space space top enclose Z subscript 2 end enclose
    rightwards double arrow   (3x + 5y) + i (5x - 3y) = -6 + 24 i
    Equating real and imaginary parts, we get
                         3x + 5y  = - 6                                                             ....(i)
    and                 5x - 3y   = 24                                                              ...(ii)
    Multiplying (i)  by 5 and (ii) by 3, we get
                 15x + 25y = -30                                                                  ...(iii)
                  15x - 9y   = 72                                                                   ...(iv)
    Subtracting (iv) from (iii), we get
                    34y  = -102                    or  space space space space space space straight y space space equals space fraction numerator negative 102 over denominator 34 end fraction equals space minus 3
    Putting y = -3 in (i), we get
              3x + 5 (-3) = -6
    or              3x = -6 +15 = 9   or      x    = 3
    Hence,      x = 3,   y = -3

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