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You probably first encountered complex numbers when you studied values of z (called roots or zeros) for which the following equation is satisfied:

a z 2 + b z + c = 0 .

For a 0 (as we will assume), this equation may be written as

z 2 + b a z + c a = 0 .

Let's denote the second-degree polynomial on the left-hand side of this equation by p ( z ) :

p ( z ) = z 2 + b a z + c a .

This is called a monic polynomial because the coefficient of the highest-power term ( z 2 ) is 1. When looking for solutions to the quadratic equation z 2 + b a z + c a = 0 , we are really looking for roots (or zeros) of the polynomial p ( z ) . The fundamental theorem of algebra says that there are two such roots. When wehave found them, we may factor the polynomial p ( z ) as follows:

p ( z ) = z 2 + b a z + c a = ( z - z 1 ) ( z - z 2 ) .

In this equation, z 1 and z 2 are the roots we seek. The factored form p ( z ) = ( z - z 1 ) ( z - z 2 ) shows clearly that p ( z 1 ) = p ( z 2 ) = 0 , meaning that the quadratic equation p ( z ) = 0 is solved for z = z 1 and z = z 2 . In the process of factoring the polynomial p ( z ) , we solve the quadratic equation and vice versa.

By equating the coefficients of z 2 , z 1 , and z 0 on the left-and right-hand sides of [link] , we find that the sum and the product of the roots z 1 and z 2 obey the equations

z 1 + z 2 = - b a z 1 z 2 = c a .

You should always check your solutions with these equations.

Completing the Square. In order to solve the quadratic equation z 2 + b a z + c a = 0 (or, equivalently, to find the roots of the polynomial z 2 + b a z + c a ) , we “complete the square” on the left-hand side of [link] :

( z + b 2 a ) 2 - ( b 2 a ) 2 + c a = 0 .

This equation may be rewritten as

( z + b 2 a ) 2 = ( 1 2 a ) 2 ( b 2 - 4 a c ) .

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 .

In the equation that defines the roots z 1 and z 2 , the term b 2 - 4 a c iscritical because it determines the nature of the solutions for z 1 and z 2 . In fact, we may define three classes of solutions depending on b 2 - 4 a c .

(i) Overdamped ( b 2 - 4 a c > 0 ) . In this case, the roots z 1 and z 2 are

z 1 , 2 = - b 2 a ± 1 2 a b 2 - 4 a c .

These two roots are real, and they are located symmetrically about the point - b 2 a . When b = 0 , they are located symmetrically about 0 at the points ± 1 2 a - 4 a c . (In this case, - 4 a c > 0 . ) Typical solutions are illustrated in [link] .

There are three cartesian graphs in this series of images. The first graph has two points on the negative portion of the x axis labeled z_2 and z_1 proceeding to the the left. In between these two points there is another point labeled -b/2a. The second graph is similar except that the point z_2 is located on the positive side of the x-axis while the point z_1 is still present at the far extreme of the negative portion of the y axis. The point -b/2a is located between these points but this time is closer to the origin. The third graph does not contain the point -b/2a, but z_1 is located on the far extreme of the negative portion of the x axis and the point z_2 is located on the far extreme of the positive x axis. There are three cartesian graphs in this series of images. The first graph has two points on the negative portion of the x axis labeled z_2 and z_1 proceeding to the the left. In between these two points there is another point labeled -b/2a. The second graph is similar except that the point z_2 is located on the positive side of the x-axis while the point z_1 is still present at the far extreme of the negative portion of the y axis. The point -b/2a is located between these points but this time is closer to the origin. The third graph does not contain the point -b/2a, but z_1 is located on the far extreme of the negative portion of the x axis and the point z_2 is located on the far extreme of the positive x axis.
Typical Roots in the Overdamped Case; (a) b / 2 a > 0 , 4 a c > 0 , (b) b / 2 a > 0 , 4 a c < 0 , and (c) b / 2 a = 0 , 4 a c < 0

(ii) Critically Damped ( b 2 - 4 a c = 0 ) . In this case, the roots z 1 and z 2 are equal (we say they are repeated):

z 1 = z 2 = - b 2 a .

These solutions are illustrated in [link] .

This Cartesian graph contains two points that rather close together. Both points exist on the negative portion of the x axis and are labeled from left to right z_1 and z_2. Above these points is the fraction -b/2a. This Cartesian graph contains two points that rather close together. Both points exist on the negative portion of the x axis and are labeled from left to right z_1 and z_2. Above these points is the fraction -b/2a. This Cartesian graph contains two points that rather close together. Both points exist on the positive portion of the x axis and are labeled from left to right z_1 and z_2. Above these points is the fraction -b/2a. This Cartesian graph contains two points that rather close together. Both points exist on the positive portion of the x axis and are labeled from left to right z_1 and z_2. Above these points is the fraction -b/2a.
Roots in the Critically Damped Case; (a) b / 2 a > 0 , and (b) b / 2 a < 0

(iii) Underdamped ( b 2 - 4 a c < 0 ) . The underdamped case is, by far, the most fascinating case. When b 2 - 4 a c < 0 , then the square root in the solutions for z 1 and z 2 ( [link] ) produces an imaginary number. We may write b 2 - 4 a c as - ( 4 a c - b 2 ) and write z 1 , 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 .

These complex roots are illustrated in [link] . Note that the roots are

purely imaginary when b = 0 , producing the result

z 1 , 2 = ± j c a .
This Cartesian graph contains a line segment that extends from the origin up and to the left into quadrant II. The ending point is labeled x and the line appears to be labeled r. An arch originates on the positive portion of the x axis and ends near the the middle of the line that was just mentioned. This arch is labeled θ. There is a point on the upper end of the positive portion of the y-axis labeled 1/2a sqrt(4ac-b^2). There is also a point in the middle of the negative portion of the x axis labeled -b/2a. Directly below this point there is a marked by an x. This Cartesian graph contains a line segment that extends from the origin up and to the left into quadrant II. The ending point is labeled x and the line appears to be labeled r. An arch originates on the positive portion of the x axis and ends near the the middle of the line that was just mentioned. This arch is labeled θ. There is a point on the upper end of the positive portion of the y-axis labeled 1/2a sqrt(4ac-b^2). There is also a point in the middle of the negative portion of the x axis labeled -b/2a. Directly below this point there is a marked by an x.  This Cartesian graph contains has a point  just above the origin on the y axis and it is labeled 1/2a sqrt(4ac-b^2). On the positive portion of the x axis there is a point. Above and below this point there is an x. At the very end of the x axis is the fraction -b/2a.  This Cartesian graph contains has a point  just above the origin on the y axis and it is labeled 1/2a sqrt(4ac-b^2). On the positive portion of the x axis there is a point. Above and below this point there is an x. At the very end of the x axis is the fraction -b/2a. This Cartesian graph has a point on the positive and negative portion of the y axis. Both points are the same distance from the origin. The upper point is labeled sqrt(c/a). This Cartesian graph has a point on the positive and negative portion of the y axis. Both points are the same distance from the origin. The upper point is labeled sqrt(c/a).
Figure 1.12: Typical Roots in the Underdamped Case; (a) b / 2 a > 0 , ( b ) b / 2 a < 0 , and (c) b / 2 a = 0

In this underdamped case, the roots z 1 and z 2 are complex conjugates:

z 2 = z 1 * .

Thus the polynomial p ( z ) = z 2 + b a z + c a = ( z - z 1 ) ( z - z 2 ) also takes the form

p ( z ) = ( z - z 1 ) ( z - z 1 * ) = z 2 - 2 Re [ z 1 ] z + | z 1 | 2 .

Re [ z 1 ] and | z 1 | 2 are related to the original coefficients of the polynomial as follows:

2 Re [ z 1 ] = - b a
| z 1 | 2 = c a

Always check these equations.

Let's explore these connections further by using the polar representations for z 1 and z 2 :

z 1 , 2 = r e ± j θ .

Then [link] for the polynomial 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 .

[link] is now

2 r cos θ = - b a r 2 = c a

These equations may be used to locate z 1 , 2 = r e ± j θ

r = c a θ = ± cos - 1 ( - b 4 a c ) .

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Source:  OpenStax, A first course in electrical and computer engineering. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col10685/1.2
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