Complex numbers: Roots and polynomials
Solving quadratic equations in ℂ
Let us return to where we began this chapter with: we have the complex unit \(\mathrm{i}\) introduced to be able to solve the equation \(z^2=-1\). But then you can solve any quadratic equation, even if the discriminant is negative. First, a few examples to illustrate:
We solve the quadratic equation by completing the square: \[\begin{aligned}z^2-6\, z+13=0 &\qquad \color{blue}{\text{the given equation}}\\ \\ \left(z-3\right)^2+4=0 &\qquad\color{blue}{\text{completing the square}}\\ \\ (z-3)^2=-4 &\qquad \color{blue}{\text{constant to the right}} \\ \\ z-3=2\,\mathrm{i}\quad \lor\quad z-3=-2\,\mathrm{i} &\qquad \color{blue}{\mathrm{i}^2=-1} \\ \\ z=3+2\,\mathrm{i}\quad \lor\quad z= 3-2\,\mathrm{i}&\qquad \color{blue}{\text{solutions in standard form}} \end{aligned}\] Above, we have used the logical "or" operator \(\lor\).
So, there are two solutions: \[z=3+2\,\mathrm{i}\quad\text{or}\quad z=3-2\,\mathrm{i}\]
So, there are two solutions: \[z=3+2\,\mathrm{i}\quad\text{or}\quad z=3-2\,\mathrm{i}\]
The following example shows that the best quadratic equation may also have complex coefficients.
Solve the following quadratic equation in \(\mathbb{C}\) by: \[z^2+2\, \ii\, z-5=0\]
We solve the quadratic equation by completing the square: \[\begin{aligned}z^2+2\, \ii\, z-5=0 &\qquad \color{blue}{\text{the given equation}}\\ \\ \left(z+\ii\right)^2-4=0 &\qquad\color{blue}{\text{completing the square with }\ii^2=-1}\\ \\ (z+\mathrm{i})^2=4 &\qquad\color{blue}{\text{constant to the right}} \\ \\z+\mathrm{i}=2\quad \lor\quad z+\mathrm{i}=-2 &\qquad\color{blue}{\text{taking roots}} \\ \\ z=2-\mathrm{i}\quad \lor\quad z=-2-\mathrm{i} &\qquad\color{blue}{\text{solutions in standard form}}\end{aligned}\] So there are two solutions: \[z=2-\mathrm{i}\quad\text{or}\quad z=-2-\mathrm{i}\]
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