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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 ) .

Questions & Answers

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Bob Reply
write examples of Nano molecule?
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The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
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Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
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Introduction about quantum dots in nanotechnology
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nano basically means 10^(-9). nanometer is a unit to measure length.
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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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