1. ## Complex Numbers Question

Hey everybody,

So I have been doing a few exercises on complex numbers, but there is one I can't seem to get out.

If x and y are real, solve the equation:

$\frac{jx}{1+jy}=\frac{3x+j4}{x+3y}$

I started manipulating it to make it more manageable:

$\frac{jx(x+3y)}{1+jy}=3x+j4$

$jx^2+j3xy=(3x+j4)(1+jy)$

etc.

I have the feeling I'm going nowhere with it. Could somebody please give me a hand?

Cheers,

Evanator

2. Originally Posted by evanator
Hey everybody,

So I have been doing a few exercises on complex numbers, but there is one I can't seem to get out.

If x and y are real, solve the equation:

$\frac{jx}{1+jy}=\frac{3x+j4}{x+3y}$

I started manipulating it to make it more manageable:

$\frac{jx(x+3y)}{1+jy}=3x+j4$

$jx^2+j3xy=(3x+j4)(1+jy)$

etc.

I have the feeling I'm going nowhere with it.
In fact, you,re going in exactly the right direction. Next step is to multiply out the brackets on the right (remembering that $j^2=-1$). Then compare the real parts and the imaginary parts on both sides of the equation.

3. $\frac{jx}{1+jy}=\frac{3x+j4}{x+3y}
$

${(jx)}{(x+3y)}=(1+jy)(3x+j4)$

$jx^2+3jxy=3x+4j+3jxy-4y$

$jx^2=3x+4j-4y$

so:

(1) $x^2=4$

(2) $0=3x-4y$

Solve this system for x and y

4. Thank you both of you. Finishing it off:

$jx^2+j3xy=3x+j3xy+j4+(j^2)4y$

$0+jx^2=(3x-4y)+j(4)$

$x^2=4$

$3x-4y=0$

$x=\pm2$

Taking the positive value of x and solving the simultaneous equations:

$3x=6$

$-3x+4y=0$

$4y=6$

$y=\frac{6}{4}=1.5$

If we take x=-2 we can see that y will come out as -1.5, so the final results are:

$x=\pm2$

$y=\pm1.5$

5. No, that is NOT the answer because x and y are not independent.

The correct answer is "x= 2 and y= 1.5 or x= -2 and y= -1.5"

Writing " $x= \pm 2, y= \pm 1.5$" implies that any combination of those, such as x= 2, y= -1.5, will satisfy the equation and that is not true.

6. Right. Thanks, HallsofIvy. I did understand that the x-value and y-value were dependent on each other, but I didn't realize I was implying otherwise with the plus-minus sign.