Complex numbers: Roots and polynomials
The quadratic-formula
We now look again at a quadratic polynomial with real coefficients. Let \(a\), \(b\), and \(c\) be real numbers with \(a\neq 0\) .
Discriminant The discriminant of the quadratic equation \(a\,z^2+b\,z+c = 0\) with real coefficients \(a\ne 0\), \(b\), and \(c\) is defined as the number \(b^2-4\cdot a \cdot c\) .
The reason for introducing the discriminant (hereafterdenoted with the letter \(D\) is that we can now formulate simply referred to) how many real solutions, the quadratic equation has and, if solutions exist, what they exactly are. The next step is to use also use the discriminant formula for complex solutions.
The formula below is called the abc-formula for complex solutions on a real kwadratisceh polynomial.
The abc-formula for complex solutions of a real polynomial The quadratic equation \(a\,z^2+b\,z+c = 0\) in unknown \(z\) and discriminant \(D=b^2-4ac\) has:
- two real solutions as \(D\gt 0\), namely \(z=\dfrac{-b - \sqrt{D}}{2a}\) and \(z=\dfrac{-b+ \sqrt{D}}{2a}\).
- exactly one real solution \(D=0\), namely \(z=-\dfrac{b}{2a}\).
- no real solutions if \(D\lt 0\).
- two complex solutions as \(D\lt 0\), namely \(z=\dfrac{-b - \sqrt{-D}\,\mathrm{i}}{2a}\) and \(z=\dfrac{-b+ \sqrt{-D}\,\mathrm{i}}{2a}\).
Proof of abc-formula (for the maths enthusiast)
Be completing the square we can rewrite the equation \(a\,z^2+b\,z+c=0\) as \[a\left(z+\frac{b}{2a}\right)^2 -\frac{b^2-4ac}{4a}=0\] Bringing the constant term to the right-hand side and dividing by \(a\), we get after some rewriting: \[\left(z+\dfrac{b}{2a}\right)^2=\dfrac{b^2-4ac}{(2a)^2}\] If \(b^2-4ac\gt 0\), then we can extract square roots on both sides; these two roots are up to a sign symbol equal to each other: \[z+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\] So there are two real solutions, namely \[z=\frac{-b - \sqrt{b^2-4ac}}{2a}\quad\mathrm{or}\quad z=\dfrac{-b+ \sqrt{b^2-4ac}}{2a}\]
If \(b^2-4ac= 0\), then the right-hand side, and also the left-hand side, is equal to \(0\), from which it follows that \[z=-\dfrac{b}{2a}\] is the only real solution.
If \(b^2-4ac\lt 0\), then the right hand side is negative, but not the left-hand side; so there are no real solutions. But there are complex solutions, namely \[z+\frac{b}{2a}=\pm\frac{\sqrt{-b^2+4ac}\,\mathrm{i}}{2a}\] So \[z=\dfrac{-b-\sqrt{-b^2+4ac}\,\mathrm{i}}{2a}\quad\mathrm{or}\quad z=\dfrac{-b+\sqrt{-b^2+4ac}\,\mathrm{i}}{2a}\]
The solutions are often taken together by making use of the \(\pm\) notation; So \[z=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\] Here we must remember that the root of a negative number is an imaginary number.
It is even better: the quadratic formula is even valid for complex numbers \(a\), \(b\) and \(c\) with \(a\neq 0\). If you must extract a square root in the formula then you may need to do it for a complex number, but you already learnt how to do this.
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The abc-formula applied in the context of complex numbers shows that each complex quadratic function has two zeros (if you count coincident zeros twice). This result can be generalized to the fundamental theorem of algebra.
Fundamental theorem of algebra Fo any polynomial \[p(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0\] of grade \(n\) there exist \(n\) complex numbers \(z_1, \ldots,z_n\) such that \[p(z)=a_n(z-z_1)\cdots(z-z_n)\] So, there are \(n\) zeros (with possible multiple-counting).
