Algebra Know-It-ALL

354 Imaginary and Complex Numbers

of−3 units is the same as a downward displacement of 3 units. This process is shown on the right

in Fig. 21-2. When we add the imaginary numbers −j3 and j2 in either order as shown, we end

up at the same point, which corresponds to −j. We have geometrically analyzed these two facts:

−j 3 +j 2 =−j

and

j 2 + (−j3)=−j

To subtract, move downward

Look again at the right-hand side of Fig. 21-2. We add a negative imaginary number to some

other imaginary number. Adding a negative imaginary number is the same thing as subtract-

ing the product of −1 and that number. If we have an imaginary number jb 1 and we want to

subtract another imaginary number jb 2 from it, we must first find the point on the number

line representing jb 1. Then we travel downward by b 2 units along the imaginary number line.

That will get us to the point representing jb 1 −jb 2.

Are you confused?

Do you suspect that the laws of real-number arithmetic apply to all imaginary numbers, and not just to j

itself? If so, you’re right! Look back at the end of Chap. 9 if you want to review those laws.

Let’s see how the distributive law, familiar with respect to multiplication and addition of real numbers,

can be used to scrutinize the two imaginary-number sums we just analyzed. We can separate the real-number

multiples, called the real coefficients, from j and then find the sums like this:

−j 3 +j 2 =j(− 3 + 2)

=j(−1)

=−j

and

j 2 + (−j3)=j[2+ (−3)]

=j(2− 3)

=j(−1)

=−j

Here’s a challenge!

In terms of the imaginary number line, express the fact that when we subtract −j5 from −j3, we get j2.

Write it down in the simplest possible form.

Solution

When we subtract a negative imaginary number, we move negatively downward along the imaginary num-

ber line, meaning that we actually travel upward. Figure 21-3 shows how this works. We start at −j3 and