absolute values equal to k. These are shown as points in Fig. C-1. There are infinitely
many complex numbers with absolute values equal to k. This fact can be shown in the
complex-number plane by plotting the set of all points at a distance of k units from the
origin. This set of points forms a circle with radius k.
Chapter 22
- To multiply out this equation, we must apply the product of sums rule on the left side
of the equals sign. To keep the signs right, let’s change the subtraction into a negative
addition before we start multiplying, and then change the negative additions back to
subtractions when we’re done. Here are the steps:
(− 7 x− 5)(− 2 x+ 9) = 0
[− 7 x+ (−5)](− 2 x+ 9) = 0
(− 7 x)× (− 2 x)+ (− 7 x)× 9 + (−5)× (− 2 x)+ (−5)× 9 = 0
14 x^2 + (− 63 x)+ 10 x+ (−45)= 0
14 x^2 + (− 53 x)+ (−45)= 0
14 x^2 − 53 x− 45 = 0
a
jb
Radius of
circle
=kunits
(0,-k)
or
- j
j
k
(0,k)
or
j
j
k
(k,0)
or
k
(-k,0)
or
- k
Figure C-1 Illustration for the solution to Prob. 10 in
Chap. 21.
Chapter 22 669