You probably first encountered complex numbers when you studied
values of zz (called roots or zeros) for which the following equation is satisfied:
az2+bz+c=0.az2+bz+c=0.
(1)
For a≠0a≠0 (as we will assume), this equation may be written as
z2+baz+ca=0.z2+baz+ca=0.(2)
Let's denote the second-degree polynomial on the left-hand side of this equation by p(z)p(z):
p(z)=z2+baz+ca.p(z)=z2+baz+ca.
(3)
This is called a monic polynomial because the coefficient of the highest-power
term (z2)(z2) is 1. When looking for solutions to the quadratic equation z2+baz+z2+baz+ca=0ca=0, we are really looking for roots (or zeros) of the polynomial p(z)p(z). The
fundamental theorem of algebra says that there are two such roots. When we
have found them, we may factor the polynomial p(z)p(z) as follows:
p(z)=z2+baz+ca=(z-z1)(z-z2).p(z)=z2+baz+ca=(z-z1)(z-z2).
(4)
In this equation, z1 and z2 are the roots we seek. The factored form p(z)=p(z)=(z-z1)(z-z2)(z-z1)(z-z2) shows clearly that p(z1)=p(z2)=0p(z1)=p(z2)=0, meaning that the
quadratic equation p(z)=0p(z)=0 is solved for z=z1z=z1 and z=z2z=z2. In the process of
factoring the polynomial p(z)p(z), we solve the quadratic equation and vice versa.
By equating the coefficients of z2,z1z2,z1, and z0 on the left-and right-hand
sides of Equation 4, we find that the sum and the product of the roots
z
1
z
1
and
z
2
z
2
obey the equations
z
1
+
z
2
=
-
b
a
z
1
z
2
=
c
a
.
z
1
+
z
2
=
-
b
a
z
1
z
2
=
c
a
.
(5)
You should always check your solutions with these equations.
Completing the Square. In order to solve the quadratic equation
z2+baz+ca=0z2+baz+ca=0 (or, equivalently, to find the roots of the polynomial z2+z2+baz+ca)baz+ca), we “complete the square” on the left-hand side of Equation 2:
(z+b2a)2-(b2a)2+ca=0.(z+b2a)2-(b2a)2+ca=0.(6)
This equation may be rewritten as
(z+b2a)2=(12a)2(b2-4ac).(z+b2a)2=(12a)2(b2-4ac).
(7)
We may take the square root of each side to find the solutions
z
1
,
2
=
-
b
2
a
±
1
2
a
b
2
-
4
a
c
.
z
1
,
2
=
-
b
2
a
±
1
2
a
b
2
-
4
a
c
.
(8)
With the roots
z
1
z
1
and
z
2
z
2
defined in Equation 1.29, prove that
(z-z1)(z-z2)(z-z1)(z-z2) is, indeed, equal to the polynomial z2+baz+caz2+baz+ca. Check that
z1+z2=-baz1+z2=-ba and z1z2=caz1z2=ca
In the equation that defines the roots
z
1
z
1
and
z
2
z
2
, the term
b
2
-
4
a
c
b
2
-
4
a
c
is
critical because it determines the nature of the solutions for
z
1
z
1
and
z
2
z
2
. In
fact, we may define three classes of solutions depending on b2-4acb2-4ac.
(i) Overdamped (b2-4ac>0)(b2-4ac>0). In this case, the roots
z
1
z
1
and
z
2
z
2
are
z1,2=-b2a±12ab2-4ac.z1,2=-b2a±12ab2-4ac.
(9)
These two roots are real, and they are located symmetrically about the point
-b2a-b2a. When b=0b=0, they are located symmetrically about 0 at the points
±12a-4ac±12a-4ac. (In this case, -4ac>0.-4ac>0.) Typical solutions are illustrated in
Figure 1.
Compute and plot the roots of the following quadratic equations:
- z2+2z+12=0z2+2z+12=0;
- z2+2a-12=0z2+2a-12=0;
- z2-12=0z2-12=0.
For each equation, check that z1+z2=-baz1+z2=-ba and z1z2=caz1z2=ca
(ii) Critically Damped (b2-4ac=0)(b2-4ac=0). In this case, the roots
z
1
z
1
and
z
2
z
2
are equal (we say they are repeated):
z1=z2=-b2a.z1=z2=-b2a.(10)
These solutions are illustrated in Figure 2.
Compute and plot the roots of the following quadratic equations:
- z2+2z+1=0z2+2z+1=0;
- z2-2z+1=0z2-2z+1=0;
- z2=0z2=0.
For each equation, check that z1+z2=-baz1+z2=-ba and z1z2=caz1z2=ca
(iii) Underdamped (b2-4ac<0)(b2-4ac<0). The underdamped case is, by
far, the most fascinating case. When b2-4ac<0b2-4ac<0, then the square root in
the solutions for
z
1
z
1
and
z
2
z
2
(Equation 8) produces an imaginary number.
We may write b2-4acb2-4ac as -(4ac-b2)-(4ac-b2) and write z1,2z1,2 as
z
1
,
2
=
-
b
2
a
±
1
2
a
-
(
4
a
c
-
b
2
)
=
-
b
2
a
±
j
1
2
a
4
a
c
-
b
2
.
z
1
,
2
=
-
b
2
a
±
1
2
a
-
(
4
a
c
-
b
2
)
=
-
b
2
a
±
j
1
2
a
4
a
c
-
b
2
.
(11)
These complex roots are illustrated in Figure 3. Note that the roots are
purely imaginary when b=0b=0, producing the result
z
1
,
2
=
±
j
c
a
.
z
1
,
2
=
±
j
c
a
.
(12)
In this underdamped case, the roots
z
1
z
1
and
z
2
z
2
are complex conjugates:
z
2
=
z
1
*
.
z
2
=
z
1
*
.
(13)
Thus the polynomial p(z)=z2+baz+ca=(z-z1)(z-z2)p(z)=z2+baz+ca=(z-z1)(z-z2) also takes the
form
p
(
z
)
=
(
z
-
z
1
)
(
z
-
z
1
*
)
=
z
2
-
2
Re
[
z
1
]
z
+
|
z
1
|
2
.
p
(
z
)
=
(
z
-
z
1
)
(
z
-
z
1
*
)
=
z
2
-
2
Re
[
z
1
]
z
+
|
z
1
|
2
.
(14)
Re [z1] Re [z1] and |z1|2|z1|2 are related to the original coefficients of the polynomial as
follows:
2
Re
[
z
1
]
=
-
b
a
2
Re
[
z
1
]
=
-
b
a
(15)
|
z
1
|
2
=
c
a
|
z
1
|
2
=
c
a
(16)
Always check these equations.
Let's explore these connections further by using the polar representations for z
1
z
1
and z
2
z
2
:
z
1
,
2
=
r
e
±
j
θ
.
z
1
,
2
=
r
e
±
j
θ
.
(17)
Then Equation 14 for the polynomial p(z)p(z) may be written in the “standard
form”
p
(
z
)
=
(
z
-
r
e
j
θ
)
(
z
-
r
e
-
j
θ
)
=
z
2
-
2
r
c
o
s
θ
z
+
r
2
.
p
(
z
)
=
(
z
-
r
e
j
θ
)
(
z
-
r
e
-
j
θ
)
=
z
2
-
2
r
c
o
s
θ
z
+
r
2
.
(18)
Equation 15 is now
2
r
cos
θ
=
-
b
a
r
2
=
c
a
2
r
cos
θ
=
-
b
a
r
2
=
c
a
(19)
These equations may be used to locate z1,2=re±jθz1,2=re±jθ
r
=
c
a
θ
=
±
cos
-
1
(
-
b
4
a
c
)
.
r
=
c
a
θ
=
±
cos
-
1
(
-
b
4
a
c
)
.
(20)
Prove that p(z)p(z) may be written as p(z)=z2-2rp(z)=z2-2r cos θz+r2θz+r2
in the underdamped case.
Prove the relations in Equation 20. Outline a graphical
procedure for locating z1=rejθz1=rejθ and z2=re-jθz2=re-jθ from the polynomial z2+z2+baz+cabaz+ca.
Compute and plot the roots of the following quadratic equations:
- z2+2z+2=0z2+2z+2=0;
- z2-2z+2=0z2-2z+2=0;
- z2+2=0z2+2=0.
For each equation, check that 2 Re [z1,2]=-ba2 Re [z1,2]=-ba and |z1,2|2=ca.|z1,2|2=ca.
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