Question A.1: Complex Arithmetic in Rectangular Form Given that Z1 = 5 + j...

For the sum, we have

Z_1+Z_2=(5+5j)+(3-j4)=8+j1

Notice that we add (algebraically) real part to real part and imaginary part to imaginary part.

The difference is

Z_1-Z_2=(5+5j)-(3-j4)=2+j9

In this case, we subtract each part of Z_2 from the corresponding part of Z_1 .

For the product, we get

\begin{matrix} Z_1Z_2&=&(5+j5)(3-j4) \\ &=&15-j20+j15-j^220 \\ &=&15-j20+j15+20 \\ &=&35-j5 \end{matrix}

Notice that we expanded the product in the usual way for binomial expressions.
Then, we used the fact that j² = –1.

To divide the numbers, we obtain

\frac{Z_1}{Z_2}=\frac{5+j5}{3-j4}

We can reduce this expression to rectangular form by multiplying the numerator and denominator by the complex conjugate of the denominator. This causes the denominator of the fraction to become pure real. Then, we divide each part of the numerator by the denominator. Thus, we find that

\begin{matrix} \frac{Z_1}{Z_2} &=&\frac{5+j5}{3-j4} \times \frac{Z_2^\ast }{Z_2^\ast } \\ &=&\frac{5+j5}{3-j4} \times \frac{3+j4}{3+j4} \\ &=&\frac{15+j20+j15+j^220}{9+j12-j12-j^216} \\ &=& \frac{15+j20+j15-20}{9+j12-j12+16} \\ &=&\frac{-5+j35}{25} \\ &=& -0.2+j1.4 \end{matrix}