360 Imaginary and Complex Numbers
The line segment connecting (3, j ) and (5, j 4) is opposite from the side with slope m 1. Let’s call its slope
m 2. Then
m 2 =Δjb/Δa
= ( j 4 −j)/(5− 3)
=j3/2
These two opposite sides are parallel. We’re halfway there! Now let’s find the slope m 3 of the line segment
connecting (2, j3) and (5, j4). It is
m 3 =Δjb/Δa
= ( j 4 −j3)/(5− 2)
=j/3
Finally, let’s find the slope m 4 of the line segment connecting (0, j 0) and (3, j ). This line segment is oppo-
site from the side with slope m 3. We have
m 4 =Δjb/Δa
= ( j− 0)/(3 − 0)
=j/3
2 4 6
j 6
j 4
j 2
a
jb
(0,j0)
(2,j3)
(5,j4)
(3,j)
Slope = m 1
Slope = m 2
Slope = m 3
Slope = m 4
Figure 21-5 Addition of (2 +j3) and (3 +j), illustrated in the
complex-number plane.