Solving equations over the complex numbers
Master solving equations over the complex numbers in Year 11 VCE Specialist Mathematics. When a real quadratic has no real solution, it still has two solutions over the complex numbers — a matched pair called a complex conjugate pair. It sits in the Algebra, number and structure area of study of the VCE Mathematics Study Design (VCAA), within the Complex numbers topic of Unit 2.
You will learn to use square-rooting, completing the square and the quadratic formula to find these roots exactly, apply the complex conjugate root theorem, and rebuild a quadratic from a single complex root — core skills for the rest of the complex numbers topic.
Theory
Solving equations over the complex numbers means finding every solution of a real quadratic in Year 11 Specialist Mathematics, even when its graph never crosses the \(x\)-axis. Over \(\mathbb{C}\) every quadratic has two solutions, and when they are not real they occur as a complex conjugate pair \(a\pm bi\).
Over the real numbers a quadratic with a negative discriminant has no solution. Over the complex numbers \(\mathbb{C}\) every quadratic has exactly two solutions, because the imaginary unit \(i\) satisfies \(i^2=-1\), so a negative number now has square roots: \(\sqrt{-k}=\sqrt{k}\,i\) for \(k>0\).
A solution is a complex number \(z=a+bi\), with real part \(a\) and imaginary part \(b\). Its complex conjugate is \(\bar{z}=a-bi\) — the same number with the sign of the imaginary part reversed.
The discriminant \(\Delta=b^2-4ac\) decides the nature of the roots. If \(\Delta<0\) the equation has two non-real solutions; if \(\Delta=0\) one repeated real solution; if \(\Delta>0\) two distinct real solutions.
The complex conjugate root theorem is the key fact: if a polynomial has real coefficients and \(a+bi\) is a root, then its conjugate \(a-bi\) is also a root. Non-real roots of a real quadratic therefore always come in conjugate pairs.
A negative number has square roots once you allow \(i\):
The quadratic formula still solves \(az^2+bz+c=0\); a negative discriminant just produces a conjugate pair:
To rebuild a monic real quadratic from a conjugate pair \(a\pm bi\), use the sum and product of its roots:
How to solve a real quadratic over \(\mathbb{C}\)
- Rearrange the equation into \(az^2+bz+c=0\) and read off \(a\), \(b\), \(c\); a quick \(\Delta=b^2-4ac\) confirms whether the roots are non-real (\(\Delta<0\)).
- Choose a method: isolate and square-root when there is no linear term, complete the square, or apply the quadratic formula.
- Simplify the negative root with \(\sqrt{-k}=\sqrt{k}\,i\), keeping surds and fractions exact.
- Write the conjugate pair \(z=a\pm bi\), and if useful check the sum \(=-\dfrac{b}{a}\) and product \(=\dfrac{c}{a}\).
Isolate \(z^2\), then take the square root of a negative number:
| \(z^2+9\) | \(=\) | \(0\) |
| \(z^2\) | \(=\) | \(-9\) |
| \(z\) | \(=\) | \(\pm\sqrt{-9}\) |
| \(=\) | \(\pm\sqrt{9}\,\sqrt{-1}\) | |
| \(=\) | \(\pm 3i\) |
\(z=\pm 3i\).
Move the constant across, then add \(\left(\tfrac{6}{2}\right)^2=9\) to both sides:
| \(z^2+6z+25\) | \(=\) | \(0\) |
| \(z^2+6z\) | \(=\) | \(-25\) |
| \(z^2+6z+9\) | \(=\) | \(-25+9\) |
| \((z+3)^2\) | \(=\) | \(-16\) |
| \(z+3\) | \(=\) | \(\pm 4i\) |
| \(z\) | \(=\) | \(-3\pm 4i\) |
\(z=-3\pm 4i\).
Apply the formula with \(a=2,\ b=-4,\ c=3\), then simplify the surd:
| \(z\) | \(=\) | \(\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) |
| \(=\) | \(\dfrac{4\pm\sqrt{(-4)^2-4(2)(3)}}{2(2)}\) | |
| \(=\) | \(\dfrac{4\pm\sqrt{16-24}}{4}\) | |
| \(=\) | \(\dfrac{4\pm\sqrt{-8}}{4}\) | |
| \(=\) | \(\dfrac{4\pm 2\sqrt{2}\,i}{4}\) | |
| \(=\) | \(1\pm\dfrac{\sqrt{2}}{2}i\) |
\(z=1\pm\dfrac{\sqrt{2}}{2}i\).
By the conjugate root theorem the conjugate is the other zero; then use \(z^2-(\text{sum})z+(\text{product})\):
| \(\text{other zero}\) | \(=\) | \(\overline{5-i}=5+i\) |
| \(\text{sum}\) | \(=\) | \((5-i)+(5+i)=10\) |
| \(\text{product}\) | \(=\) | \((5-i)(5+i)\) |
| \(=\) | \(25-i^2\) | |
| \(=\) | \(25+1\) | |
| \(=\) | \(26\) | |
| \(z^2-(\text{sum})z+\text{product}\) | \(=\) | \(z^2-10z+26\) |
\(z^2-10z+26\).
Common pitfalls
Frequently asked questions
Can every quadratic be solved over the complex numbers?
Yes. Over \(\mathbb{C}\) every quadratic has exactly two solutions (a repeated root counts twice). When the discriminant is negative the two solutions are a non-real conjugate pair.
What does the discriminant tell you about complex roots?
If \(\Delta=b^2-4ac<0\) the quadratic has two complex conjugate roots \(a\pm bi\); if \(\Delta=0\) a repeated real root; if \(\Delta>0\) two distinct real roots.
What is the complex conjugate root theorem?
If a polynomial has real coefficients and \(a+bi\) is a root, then \(a-bi\) is also a root. So non-real roots of a real quadratic always come in conjugate pairs.
How do you take the square root of a negative number?
Use \(\sqrt{-k}=\sqrt{k}\,i\) for \(k>0\). For example \(\sqrt{-9}=3i\) and \(\sqrt{-8}=2\sqrt{2}\,i\); keep surds exact.
How do you find a real quadratic given one complex root?
The conjugate is the other root. Build \(z^2-(\text{sum})z+(\text{product})\); for \(a\pm bi\) the sum is \(2a\) and the product is \(a^2+b^2\).
Do complex solutions appear on the normal \(x\)-\(y\) graph?
No. If the roots are non-real the parabola never crosses the \(x\)-axis. The roots are shown instead on an Argand (complex) plane, symmetric about the real axis.