To solve the given equation:
[tex]\[
\frac{j x}{1 + i y} = \frac{3 x + j 4}{x + 3 y}
\][/tex]
where [tex]\(j = \sqrt{-1}\)[/tex], we will proceed through the steps to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
1. Equate the imaginary and real parts: From the given equation, we see that both sides must have the same real and imaginary parts.
Since [tex]\(\frac{j x}{1 + i y}\)[/tex] should be equal to [tex]\(\frac{3 x + j 4}{x + 3 y}\)[/tex], separate real and imaginary parts on both sides.
2. Handle the denominators: To get rid of the denominators, cross-multiply:
[tex]\[
(j x)(x + 3 y) = (3 x + j 4)(1 + i y)
\][/tex]
3. Expand both sides: Multiply out terms on both sides:
[tex]\[
j x \cdot x + j x \cdot 3 y = (3 x \cdot 1 + 3 x \cdot i y) + (j 4 \cdot 1 + j 4 \cdot i y)
\][/tex]
[tex]\[
j x^2 + j 3 x y = 3 x + 3 x i y + j 4 + j 4 i y
\][/tex]
4. Separate real and imaginary parts:
[tex]\[
j x^2 + j 3 x y = 3 x + 3 x i y + j 4 + j 4 i y
\][/tex]
5. Combine like terms: Collect real and imaginary components on each side and equate them:
Real Part:
[tex]\[
3 x = 0
\][/tex]
Imaginary Part:
[tex]\[
j x^2 + j 3 x y = j 4 + j 4 i y + 3 x i y
\][/tex]
From the real part equation:
[tex]\[
3 x = 0 \implies x = 0 \quad \text{(not useful, move to imaginary parts)}
\][/tex]
6. Simultaneously solve the system:
[tex]\[
x^2 + 3 x y = 4 + 4 y \quad \implies x^2 + 3 x y - 4 - 4 y = 0
\][/tex]
And since [tex]\(j x = 3x j \implies x = 3 \implies y = -\frac{1}{4}\)[/tex], but we rather rely on:
By solving simultaneously for both:
The resulting pairs of solutions to the equation [tex]\( \frac{jx}{1+iy}=\frac{3 x+ j4}{x+3y} \)[/tex] are:
[tex]\[
\boxed{(-2, -3/2) \quad \text{and} \quad (2, 3/2)}
\][/tex]
These solutions respect the separables for real numbers and imaginary correspondence.
