if x +y are real solve the equation jx/1+jy = 3x +j4/x +3y

complex numbers help please

if x +y are real solve the equation

jx/1+jy = 3x +j4/x +3y
in Algebra 2 Answers by

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1 Answer

Assume:j=√-1 and jx/(1+jy)=(3x+j4)/(x+3y); cross multiply:

jx(x+3y)=(3x+j4)(1+jy)=3x+3jxy+j4-4y,

jx2+3jxy=3x+3jxy+j4-4y,

jx2=3x+j4-4y; separate real and imaginary components:

Imaginary:

x2=4, x=±2;

Real:

3x-4y=0, 3x=4y, y=¾x, so y=±3/2.

SOLUTION: (2,3/2) or (-2,-3/2) for (x,y).

by Top Rated User (1.1m points)

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