In the bottom equation, we can add the quantity (2 +j3) to each side, getting
x= 2 +j 3
The roots can be formally expressed this way:
x=− 2 −j3 or x= 2 +j 3
The solution set is
X= {(− 2 −j3), (2 +j3)}
- To get the polynomial form of the quadratic stated in Prob. 7, we can multiply out the
trinomial factors. Here’s the original equation again:
(x+ 2 +j3)(x− 2 −j3)= 0
Let’s convert the subtractions in the second factor to negative additions individually to
minimize the risk of getting the signs mixed up when we expand the equation into poly-
nomial form. That gives us
(x+ 2 +j3)[x+ (−2)+ (−j3)]= 0
Now we can multiply out, obtaining
x^2 + (− 2 x)+ (−j 3 x)
+ 2 x+ (−4)+ (−j6)
+j 3 x+ (−j6)+ 9
= 0
which simplifies to
x^2 + (−j12)+ 5 = 0
and further to
x^2 + (5 −j12)= 0
- Here’s the polynomial equation we derived. It’s interesting, because the coefficient of x
is equal to 0, while the stand-alone constant is complex.
x^2 + (5 −j12)= 0
Here are the roots we found:
x=− 2 −j3 or x= 2 +j 3
Chapter 23 679
