Given z = (a+jb)/(c+jd), where a, b, c, d are real quantities, show that if z is real, then (a)/(b) = (c)/(d), and if z is entirely imaginary, then ac = -bd.

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

Since $j^2 = -1$:

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

Group the real and imaginary terms:

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

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

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

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

Assuming $b \neq 0$ and $d \neq 0$, divide both sides by $bd$:

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

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

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

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

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

Assuming $b \neq 0$ and $c \neq 0$, divide both sides by $bc$:

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

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

First, expand $(1+j)^2$:

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

Next, expand $j(2+j)$:

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

Now, add these two results:

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

The original equation becomes:

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

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