Solving equations over the complex numbers
Learn how to solve equations over the complex numbers in Year 11 Specialist Mathematics for Queensland (QCAA). 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.
You will 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 (QCAA, Queensland), 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.