Complexities

Algebra Level 3

Suppose x x and y y are real numbers and 4 x + 3 x i + y i 43 = 2 y + 2 x i x + 2 y i + 5 i 4x+3xi+yi-43 = 2y+2xi-x+2yi+5i . Find the product x y xy .

Details and assumptions

i i is the imaginary unit, where i 2 = 1 i^2=-1 .


The answer is 66.

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1 solution

Arron Kau Staff
May 13, 2014

Equating the real and imaginary parts of the complex numbers gives us a system of linear equations to solve: { 4 x 43 = 2 y x 3 x + y = 2 x + 2 y + 5 \begin{cases} 4x - 43 = 2y-x \\ 3x+y = 2x+2y+5 \\ \end{cases} Solving the first equation for y y gives us y = 5 x 43 2 y = \frac{5x-43}{2} . Substituting this expression in for y y in the second equation yields 3 x + 5 x 43 2 = 2 x + 5 x 43 + 5 3x + \frac{5x-43}{2} = 2x + 5x - 43 + 5 , and multiplying both sides by 2 2 to get rid of the fraction gives us 6 x + 5 x 43 = 4 x + 10 x 86 + 10 6x+5x-43 = 4x + 10x - 86 + 10 . This equation simplifies to 33 = 3 x 33 = 3x , from which it follows that x = 11 x = 11 . Hence, y = 55 43 2 = 6 y = \frac{55 - 43}{2} = 6 , and the product x y xy must be 66.

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