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25 November, 11:59

Which values of x and y would make the following expression represent a real number?

(6 + 3i) (x + yi)

x = 6, y = 0

x = - 3, y = 0

x = 6, y = - 3

x = 0, y = - 3

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Answers (1)
  1. 25 November, 12:25
    0
    this expression represent a real number when the imaginer part = 0

    (6+3i) (x+yi) = 6x+6yi+3ix+3yi² ... i²=-1

    (6+3i) (x+yi) = 6x+6yi+3ix-3y

    (6+3i) (x+yi) = 6x-3y + (6y+3x) i ... standard form

    the imaginer part is : 6y+3x=0

    for x=6 and y = - 3 : 6 (-3) + 3 (6) = 0
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