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Complex Numbers: Roots of Quadratic Equations

Module by: Louis Scharf. E-mail the author

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 a0a0 (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)

Exercise 1

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.

Figure 1: Typical Roots in the Overdamped Case; (a) b/2a>0,4ac>0b/2a>0,4ac>0, (b) b/2a>0,4ac<0b/2a>0,4ac<0, and (c) b/2a=0,4ac<0b/2a=0,4ac<0
(a) (b) (c)
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.Figure 1(b) (pic010.png)Figure 1(c) (pic011.png)

Exercise 2

Compute and plot the roots of the following quadratic equations:

  1. z2+2z+12=0z2+2z+12=0;
  2. z2+2a-12=0z2+2a-12=0;
  3. 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.

Figure 2: Roots in the Critically Damped Case; (a) b/2a>0b/2a>0, and (b) b/2a<0b/2a<0
(a) (b)
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.

Exercise 3

Compute and plot the roots of the following quadratic equations:

  1. z2+2z+1=0z2+2z+1=0;
  2. z2-2z+1=0z2-2z+1=0;
  3. 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)
Figure 3: Figure 1.12: Typical Roots in the Underdamped Case; (a) b/2a>0,(b)b/2a>0,(b)b/2a<0b/2a<0, and (c) b/2a=0b/2a=0
(a) (b) (c)
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 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).

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)

Exercise 4

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.

Exercise 5

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.

Exercise 6

Compute and plot the roots of the following quadratic equations:

  1. z2+2z+2=0z2+2z+2=0;
  2. z2-2z+2=0z2-2z+2=0;
  3. 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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