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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:
For $a\ne 0$ (as we will assume), this equation may be written as
Let's denote the second-degree polynomial on the left-hand side of this equation by $p\left(z\right)$ :
This is called a monic polynomial because the coefficient of the highest-power term $\left({z}^{2}\right)$ is 1. When looking for solutions to the quadratic equation ${z}^{2}+\frac{b}{a}z+$ $\frac{c}{a}=0$ , we are really looking for roots (or zeros) of the polynomial $p\left(z\right)$ . The fundamental theorem of algebra says that there are two such roots. When wehave found them, we may factor the polynomial $p\left(z\right)$ as follows:
In this equation, z _{1} and z _{2} are the roots we seek. The factored form $p\left(z\right)=$ $(z-{z}_{1})(z-{z}_{2})$ shows clearly that $p\left({z}_{1}\right)=p\left({z}_{2}\right)=0$ , meaning that the quadratic equation $p\left(z\right)=0$ is solved for $z={z}_{1}$ and $z={z}_{2}$ . In the process of factoring the polynomial $p\left(z\right)$ , 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
You should always check your solutions with these equations.
Completing the Square. In order to solve the quadratic equation ${z}^{2}+\frac{b}{a}z+\frac{c}{a}=0$ (or, equivalently, to find the roots of the polynomial ${z}^{2}+$ $\frac{b}{a}z+\frac{c}{a})$ , we “complete the square” on the left-hand side of [link] :
This equation may be rewritten as
We may take the square root of each side to find the solutions
In the equation that defines the roots ${z}_{1}$ and ${z}_{2}$ , the term ${b}^{2}-4ac$ 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 ${b}^{2}-4ac$ .
(i) Overdamped $({b}^{2}-4ac>0)$ . In this case, the roots ${z}_{1}$ and ${z}_{2}$ are
These two roots are real, and they are located symmetrically about the point $-\frac{b}{2a}$ . When $b=0$ , they are located symmetrically about 0 at the points $\pm \frac{1}{2a}\sqrt{-4ac}$ . (In this case, $-4ac>0.$ ) Typical solutions are illustrated in [link] .
(ii) Critically Damped $({b}^{2}-4ac=0)$ . In this case, the roots ${z}_{1}$ and ${z}_{2}$ are equal (we say they are repeated):
These solutions are illustrated in [link] .
(iii) Underdamped $({b}^{2}-4ac<0)$ . The underdamped case is, by far, the most fascinating case. When ${b}^{2}-4ac<0$ , then the square root in the solutions for ${z}_{1}$ and ${z}_{2}$ ( [link] ) produces an imaginary number. We may write ${b}^{2}-4ac$ as $-(4ac-{b}^{2})$ and write ${z}_{1,2}$ as
These complex roots are illustrated in [link] . Note that the roots are
purely imaginary when $b=0$ , producing the result
In this underdamped case, the roots ${z}_{1}$ and ${z}_{2}$ are complex conjugates:
Thus the polynomial $p\left(z\right)={z}^{2}+\frac{b}{a}z+\frac{c}{a}=(z-{z}_{1})(z-{z}_{2})$ also takes the form
$Re\left[{z}_{1}\right]$ and $|{z}_{1}{|}^{2}$ are related to the original coefficients of the polynomial as follows:
Always check these equations.
Let's explore these connections further by using the polar representations for ${z}_{1}$ and ${z}_{2}$ :
Then [link] for the polynomial $p\left(z\right)$ may be written in the “standard form”
[link] is now
These equations may be used to locate ${z}_{1,2}=r{e}^{\pm j\theta}$
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