F.4 Maths_1－ppi93gk88d的部落格

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F.4 Maths

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If x and y are real numbers and (5-3i)/(4+3i) = (x+y) +(3x-y)i find thevalue of x and y.

最佳解答:

(5 - 3i)/(4 + 3i) = (x + y) + (3x - y)i Consider LHS = (5 - 3i)/(4 + 3i) = (5 - 3i)/(4 + 3i) × (4 - 3i)/(4 - 3i) = [(5 - 3i) × (4 - 3i)] / [(4 + 3i) × (4 - 3i)] = (20 - 15i - 12i + 9i2) / [42 - (3i)2] = (20 - 27i - 9) / (16 + 9) = (11 - 27i) / 25 = 11/25 - (27/25)i Therefore, consider { x + y = 11/25 { 3x - y = -27/25 4x = -16/25 x = -4/25 y = 11/25 - x = 15/25 = 3/5 The answer is: x = -4/25 y = 3/5 2014-09-14 19:14:18 補充：噢，Leo，謝謝你～你真有風度！加油！ ╭∧---∧╮ │ .??? │ ╰/) ? (\╯

其他解答:

(5 - 3i) / (4 + 3i) = (x + y) + (3x - y)i (5 - 3i) = (4 + 3i)[(x + y) + (3x - y)i] (5 - 3i) = 4x + 4y + 12xi - 4yi + 3xi + 3yi - 9x + 3y (5 - 3i) = (-5x + 7y) + (15x - y)i So, -5x + 7y = 5, 15x - y = -3 y = 15x + 3 -5x + 7(15x + 3) = 5 100x = -16 x = -4/25 y = 15x + 3 = 3/5|||||迟咗几秒.搬到意見柵 (5-3i)/(4+3i) = (x+y) +(3x-y)i (5-3i) = (4+3i)[(x+y) +(3x-y)i] (5-3i) = [4x+4y +(12x-4y)i+(3x+3y)i +(9x-3y)i^2] (5-3i) = [4x+4y +(15x-y)i -(9x-3y)] (5-3i) = 7y-5x +(15x-y)i ∴ 5=7y-5x ___(1) -3=15x-y___(2) (1)*3+(2), 12=20y y=3/5 put y=3/5 into (2) -3=15x-3/5 x=-4/25 ∴ x=-4/25 y=3/5