If x and y are real, solve the equation: (jx)/(1+jy) = (3x+j4)/(x+3y)

Remember that $j^2 = -1$.

$$jx^2 + j3xy = 3x + j3xy + j4 + j^24y$$ $$jx^2 + j3xy = 3x + j3xy + j4 - 4y$$

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

Since $x$ and $y$ are real, the left side is purely imaginary and the right side is purely real. For them to be equal, both must be zero.
Equating the real parts:

$$3x - 4y = 0 \quad (*)$$

Equating the imaginary parts:

$$x^2 - 4 = 0 \quad (**)$$

$$x^2 = 4$$ $$x = \pm 2$$

Case 1: $x=2$

$$3(2) - 4y = 0$$ $$6 - 4y = 0$$ $$4y = 6$$ $$y = \frac{6}{4} = \frac{3}{2}$$

Case 2: $x=-2$

$$3(-2) - 4y = 0$$ $$-6 - 4y = 0$$ $$4y = -6$$ $$y = \frac{-6}{4} = -\frac{3}{2}$$

The solutions are $\boxed{x=2, y=\frac{3}{2}}$ and $\boxed{x=-2, y=-\frac{3}{2}}$.

Question 11:
If $z = \frac{a+jb}{c+jd}$, where $a, b, c$ and $d$ are real quantities, show that

(a) if $z$ is real then $\frac{a}{b} = \frac{c}{d}$

(b) if $z$ is entirely imaginary then $\frac{a}{b} = -\frac{d}{c}$

$$z = \frac{a+jb}{c+jd} \times \frac{c-jd}{c-jd}$$ $$z = \frac{(a+jb)(c-jd)}{c^2 - (jd)^2}$$ $$z = \frac{ac - ajd + jbc - j^2bd}{c^2 - j^2d^2}$$

Since $j^2 = -1$:

$$z = \frac{ac - ajd + jbc + bd}{c^2 + d^2}$$ $$z = \frac{(ac+bd) + j(bc-ad)}{c^2+d^2}$$ $$z = \frac{ac+bd}{c^2+d^2} + j\frac{bc-ad}{c^2+d^2}$$

Let the real part be $X = \frac{ac+bd}{c^2+d^2}$ and the imaginary part be $Y = \frac{bc-ad}{c^2+d^2}$.

a) If $z$ is real, its imaginary part must be zero.

$$Y = 0$$ $$\frac{bc-ad}{c^2+d^2} = 0$$

Since $c^2+d^2 \neq 0$, the numerator must be zero:

$$bc-ad = 0$$ $$bc = ad$$

Divide both sides by $bd$ (assuming $b \neq 0$ and $d \neq 0$):

$$\frac{bc}{bd} = \frac{ad}{bd}$$ $$\frac{c}{d} = \frac{a}{b}$$

This shows that if $z$ is real, then $\boxed{\frac{a}{b} = \frac{c}{d}}$.

b) If $z$ is entirely imaginary, its real part must be zero.

$$X = 0$$ $$\frac{ac+bd}{c^2+d^2} = 0$$

Since $c^2+d^2 \neq 0$, the numerator must be zero:

$$ac+bd = 0$$ $$ac = -bd$$

Divide both sides by $bc$ (assuming $b \neq 0$ and $c \neq 0$):

$$\frac{ac}{bc} = \frac{-bd}{bc}$$ $$\frac{a}{b} = -\frac{d}{c}$$

This shows that if $z$ is entirely imaginary, then $\boxed{\frac{a}{b} = -\frac{d}{c}}$.

Question 12:
Given that $(a+b)+j(a-b) = (1+j)^2 + j(2+j)$, obtain the values of $a$ and $b$.

$$(1+j)^2 + j(2+j)$$

Expand $(1+j)^2$:

$$(1+j)^2 = 1^2 + 2(1)(j) + j^2 = 1 + 2j - 1 = 2j$$

Expand $j(2+j)$:

$$j(2+j) = 2j + j^2 = 2j - 1$$

Now, add the simplified parts:

$$2j + (2j - 1) = 4j - 1$$

So, the RHS is $-1 + 4j$.

The equation is now:

$$(a+b)+j(a-b) = -1 + 4j$$

Equating the real parts:

$$a+b = -1 \quad (1)$$

Equating the imaginary parts:

$$a-b = 4 \quad (2)$$

Add equation (1) and equation (2):

$$(a+b) + (a-b) = -1 + 4$$ $$2a = 3$$ $$a = \frac{3}{2}$$

Substitute the value of $a$ into equation (1):

$$\frac{3}{2} + b = -1$$ $$b = -1 - \frac{3}{2}$$ $$b = -\frac{2}{2} - \frac{3}{2}$$ $$b = -\frac{5}{2}$$

The values are $\boxed{a=\frac{3}{2}, b=-\frac{5}{2}}$.