Here's a problem that's similar in spirit to problem number 2 on the review sheet but just a bit harder:
Use the definition of complex numbers as the set of ordered pairs of real numbers and the associated definitions of addition and multiplication to show that complex multiplication is distributive over complex addition.
Right so I think I have an answer for this one.
First definitions. Let
$Z_1,Z_2,Z_3\in \mathbb{C}$, where $Z_1=(x_1,y_1)$, $Z_2=(x_2,y_2)$,
and $Z_3=(x_3,y_3)$,
with $x_1,x_2,x_3,y_1,y_2,y_3 \in \mathbb{R}$.
Then, we want to prove that $$Z_1(Z_2+Z_3)=Z_1Z_2+Z_1Z_3.$$
Thus
\begin{align*}
Z_1(Z_2+Z_3)&=(x_1,y_1)[(x_2,y_2)+(x_3,y_3)]\\
&=(x_1,y_1)(x_2+x_3,y_2+y_3)\\
&=(x_1(x_2+x_3)-y_1(y_2+y_3),x_1(y_2+y_3)+y_1(x_2+x_3))\\
&=(x_1x_2+x_1x_3-y_1y_2-y_1y_3,x_1y_2+x_1y_3+y_1x_2+y_1x_3)\\
&=((x_1x_2-y_1y_2)+(x_1x_3-y_1y_3),(x_1y_2+y_1x_2)+(x_1y_3+y_1x_3))\\
&=(x_1x_2-y_1y_2,x_1y_2+y_1x_2)+(x_1x_3-y_1y_3,x_1y_3+y_1x_3)\\
&=(x_1,y_1)(x_2,y_2)+(x_1,y_1)(x_3,y_3)\\
&=Z_1 Z_2 + Z_1 Z_3
\end{align*}
End