Ultimate Guide to Solving Equations with Complex Numbers

Introduction

At the very beginning of our complex numbers journey, we used the imaginary unit $i$ to solve quadratic equations with negative discriminants, producing roots like $x = 2 \pm 3i$ . But what if the equation itself contains imaginary numbers in its coefficients? What if the polynomial has a degree of 3, 4, or 5? In this guide, Solving Equations with Complex Numbers, we explore advanced root-finding techniques and the elegant symmetry of the Complex Conjugate Root Theorem.

Executive Summary

This guide connects complex numbers with polynomial theory:

Quadratics with Complex Coefficients: Solving $az^2 + bz + c = 0$ where $a, b, c$ can be complex numbers, using the quadratic formula and complex square roots.
The Complex Conjugate Root Theorem: Understanding why non-real roots must always appear in conjugate pairs (e.g., $a+ib$ and $a-ib$ ) when the polynomial has real coefficients.
Factoring Polynomials: Using conjugate pairs to construct real quadratic factors and break down high-degree polynomials.

What is this about?

The Fundamental Theorem of Algebra states that a polynomial of degree $n$ will have exactly $n$ roots over the complex field. This means a cubic will always have 3 roots, and a quartic will always have 4 roots.

However, if you graph a cubic polynomial with real coefficients, you might only see it cross the x-axis once. Where are the other two roots? They exist, but they are complex, floating off the real number line. This guide teaches you how to find them, and more importantly, how to use the predictable "conjugate pair" pattern to reconstruct the original equation.

Main Content

1. Solving Quadratics with Complex Coefficients

Consider the equation $z^2 + (2+i)z + (3-i) = 0$ . Here, the coefficients are complex numbers. The standard quadratic formula still works: $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The challenge arises when you calculate the discriminant $\Delta = b^2 - 4ac$ . The result will usually be a complex number. To evaluate $\sqrt{\Delta}$ , you cannot use a calculator. You must use the simultaneous equation method you learned in Complex Arithmetic (Part 3) to find the two complex square roots of $\Delta$ .

Once you find the square roots, you substitute them back into the $\pm \sqrt{\Delta}$ part of the quadratic formula to find the two values for $z$ .

Note: The roots of quadratics with complex coefficients will not be conjugate pairs.

2. The Complex Conjugate Root Theorem

If a polynomial equation $P(z) = 0$ has strictly real coefficients, and you are told that $z = a + ib$ is a root, then the theorem guarantees that its complex conjugate, $\bar{z} = a - ib$ , must also be a root.

Why? The proof relies on the conjugate properties we learned earlier: If $P(z) = c_n z^n + \dots + c_1 z + c_0 = 0$ , where all $c_k$ are real. Take the conjugate of both sides: $\overline{c_n z^n + \dots + c_1 z + c_0} = \bar{0}$ Since the conjugate of a sum is the sum of conjugates, and the conjugate of a product is the product of conjugates: $\overline{c_n}\overline{z^n} + \dots + \overline{c_1}\bar{z} + \overline{c_0} = 0$ Since the coefficients $c_k$ are real, their conjugates are just themselves ( $\overline{c_k} = c_k$ ): $c_n(\bar{z})^n + \dots + c_1(\bar{z}) + c_0 = 0$ This is exactly the statement that $P(\bar{z}) = 0$ . Thus, $\bar{z}$ is a root!

3. Using Conjugate Roots to Factor Polynomials

If you know that $z = a + ib$ and $z = a - ib$ are roots of a polynomial with real coefficients, you know that $(z - (a+ib))$ and $(z - (a-ib))$ are linear factors.

If you multiply these two linear factors together: $(z - (a+ib))(z - (a-ib)) = (z-a - ib)(z-a + ib)$ This is a difference of two squares: $= (z-a)^2 - (ib)^2$ $= z^2 - 2az + a^2 + b^2$

Notice that all the $i$ 's have vanished! The result is a quadratic factor with purely real coefficients. You can then use polynomial long division to divide the original high-degree polynomial by this real quadratic factor to find the remaining roots.

Simple Worked Example

Question: The equation $z^3 - 3z^2 + 9z + 13 = 0$ has real coefficients. One of its roots is $z = 2 + 3i$ . Find the other two roots.

Solution: Step 1: Apply the Complex Conjugate Root Theorem Since the coefficients ( $1, -3, 9, 13$ ) are all real, the complex roots must appear in conjugate pairs. Therefore, if $z_1 = 2 + 3i$ is a root, then $z_2 = 2 - 3i$ is also a root.

Step 2: Construct the quadratic factor The linear factors are $(z - (2+3i))$ and $(z - (2-3i))$ . Multiply them together: $= ((z-2) - 3i)((z-2) + 3i)$ $= (z-2)^2 - (3i)^2$ $= (z^2 - 4z + 4) - (-9)$ $= z^2 - 4z + 13$

This quadratic is a factor of the original cubic.

Step 3: Find the third root Let the third root be $\alpha$ (which must be a real number, since the other two are a conjugate pair). The cubic can be factored as: $z^3 - 3z^2 + 9z + 13 = (z^2 - 4z + 13)(z - \alpha)$

We can find $\alpha$ quickly by comparing the constant terms on both sides: Constant on left: $13$ Constant on right: $13 \times (-\alpha) = -13\alpha$ $13 = -13\alpha \implies \alpha = -1$

Therefore, the third root is $z_3 = -1$ . (You could also find this using polynomial long division).

Answer: The other two roots are $2 - 3i$ and $-1$ .

mini-FAQ page

Q: What if the polynomial has complex coefficients? Does the conjugate theorem still apply? A: No! The Complex Conjugate Root Theorem strictly requires the polynomial to have purely real coefficients. If you have an equation like $z^2 + iz + 2 = 0$ , the roots will not be conjugates of each other.

Q: Can a cubic with real coefficients have three complex roots? A: No. Complex roots must come in conjugate pairs (groups of 2). A cubic has 3 roots total. Therefore, it can have either 0 complex roots (and 3 real) or 2 complex roots (and 1 real). A polynomial with real coefficients and an odd degree must always have at least one real root.

Common mistakes to avoid

Applying the theorem when coefficients are not real: Always check the coefficients before assuming a conjugate root exists. If a question gives you $P(z) = z^3 + (1+i)z + 4$ , the theorem does not apply.
Algebraic errors in quadratic construction: When multiplying $(z - (a+ib))(z - (a-ib))$ , students often mess up the signs. Remember to group the real parts $(z-a)$ together first to create the difference of two squares easily.

Practice on Vu's Maths Hub

Mastering polynomial root theorems is a massive mark-earner in the HSC.

Practice polynomial long division and conjugate pair factoring in the HSC Polynomials Booklet.
Solve brutal quadratics with complex coefficients in the HSC Complex Numbers Booklet.

Connect with me

Want to master Complex Polynomials and guarantee top marks in HSC Mathematics Extension 2? Visit Vu's Maths Hub for in-depth booklets, rigorous worked solutions, and expert advice to help you ace your exams!