Exercise 65 Page 109

When we look at the expression, we see that there have been given sums, differences and products of the complex numbers. To add or subtract two complex numbers, we add or subtract their real parts and imaginary parts. When we multiply two complex numbers, we use Distributive Property and we have $i^2$ as a result of this operation. In this case, we should simplify $i^2$ as $\text{-}1.$ Keeping these in mind, let's simplify each of the corresponding expressions.

Given	Simplified
$(\text{-}4+7i)+(\text{-}4-7i)$	$\text{-}8+0i$
$(2-6i)-(\text{-}10+4i)$	$12-10i$
$(25+15i)-(25-6i)$	$0+21i$
$(5+i)(8-i)$	$41+3i$
$(17-3i)+(\text{-}17-6i)$	$0-9i$
$(\text{-}1+2i)(11-i)$	$\text{-}9+23i$
$(7+5i)+(7-5i)$	$14+0i$
$(\text{-}3+6i)-(\text{-}3-8i)$	$0+14i$

Now, the expressions are written in standard form. A complex number written in standard form is a number $a+bi$ where $a$ and $b$ are real numbers. The number $a$ is the real part, and the number $bi$ is the imaginary part. Therefore, we have three cases.

• If $b\ne 0,$ then $a+bi$ is an imaginary number.
• If $b=0,$ then $a+bi$ is a real number.
• If $a=0,$ then $a+bi$ is a pure imaginary number.

With these, let's classify our results.

Real numbers	Imaginary numbers	Pure imaginary numbers
$$\begin{array}{c}\text{-}8+0i\\ 14+0i\end{array}$$	$$\begin{array}{c}12-10i\\ 41+3i\\ \text{-}9+23i\end{array}$$	$$\begin{array}{c}0+21i\\ 0-9i\\ 0+14i\end{array}$$