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

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

$$x=-2, y=-1.5$$

$$x=2, y=1.5$$

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