Given z = (a+jb)/(c+jd), where a, b, c, d are real quantities. a) Show that if z is real, then (a)/(b) = (c)/(d)

Okay, Hølüwårh, let's go through Question 11 again.

Given $z = \frac{a+jb}{c+jd}$, where $a, b, c, d$ are real quantities.

Step 1: Express $z$ in the form $X+jY$ by multiplying the numerator and denominator by the conjugate of the denominator.
The conjugate of the denominator $c+jd$ is $c-jd$.

$$z = \frac{a+jb}{c+jd} \times \frac{c-jd}{c-jd}$$

Multiply the numerators and denominators:

$$z = \frac{(a+jb)(c-jd)}{c^2 - (jd)^2}$$

Expand the numerator: $(a+jb)(c-jd) = ac - ajd + jbc - j^2bd$.
Simplify the denominator: $c^2 - (jd)^2 = c^2 - j^2d^2$.
Since $j^2 = -1$, substitute this into both numerator and denominator:

$$z = \frac{ac - ajd + jbc - (-1)bd}{c^2 - (-1)d^2}$$ $$z = \frac{ac - ajd + jbc + bd}{c^2 + d^2}$$

Group the real and imaginary parts in the numerator:

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

Separate into real and imaginary components:

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

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