Answer to Question #254427 in Complex Analysis for trung

Question

If x and y are real, solve the equation
jx/1+jy=3x+4j/x+3y

Answer 1

$\dfrac{jx}{1+jy}=\dfrac{3x+4j}{x+3y}$

$j(x^2+xy)=3x-4y+j(xy+4)$

If $x$ and $y$ are real

$\begin{cases} 3x-4y=0 \\ x^2+xy=xy+4 \end{cases}$

$x=-2, y=-1.5$

$x=2, y=1.5$

$(-2, -1.5),\ (2,1.5)$