Complex numbers

Q: What is the definition of the imaginary unit \(i\) and how is it used to simplify square roots of negative numbers?

The imaginary unit is defined by \(i = \sqrt{-1}\), which means \(i^2 = -1\). Square roots of negative numbers simplify as \(\sqrt{-k} = \sqrt{k}i\) for \(k > 0\).

Q: What is the modulus-argument form of a complex number and how do you multiply and divide numbers in this form?

Modulus-argument form is \(z = r(\cos\theta + i\sin\theta)\). When multiplying, multiply moduli and add arguments; when dividing, divide moduli and subtract arguments.

Further Mathematics 2330 · CCEA AS Level · 9 min read · Updated: 31 Aug 2026

Chapter: Complex Numbers (CCEA AS 1: Pure Mathematics)

Welcome to Complex Numbers! Throughout GCSE and standard AS Mathematics, whenever you tried to take the square root of a negative number like \(\sqrt{-16}\), your calculator gave an error, and you were told "there are no real solutions."

In Further Mathematics, we expand our number system. Complex numbers unlock solutions to equations that were previously impossible to solve, providing essential mathematical tools for physics, electronics, aerodynamics, and computer graphics. Don't worry if this feels a bit abstract at first—we will break down every single rule step-by-step.

1. The Imaginary Unit and Cartesian Form

To deal with the square roots of negative numbers, mathematicians defined the imaginary unit, denoted by the letter \(i\).

Definition of \(i\)

\(i = \sqrt{-1}\) which means \(i^2 = -1\)

Using this rule, we can simplify any negative square root:
\(\sqrt{-25} = \sqrt{25 \times (-1)} = \sqrt{25} \times \sqrt{-1} = 5i\)
\(\sqrt{-7} = \sqrt{7 \times (-1)} = \sqrt{7}i\)

Cartesian Form of a Complex Number

A complex number \(z\) is usually written in Cartesian form (also called standard form):

\(z = x + iy\) (or \(a + ib\)), where \(x, y \in \mathbb{R}\)

Real part: \(\text{Re}(z) = x\)
Imaginary part: \(\text{Im}(z) = y\)

Crucial Examiner Warning: Notice that the imaginary part \(\text{Im}(z)\) is strictly the real number \(y\), not \(yi\).
Example: If \(z = 4 - 3i\), then \(\text{Re}(z) = 4\) and \(\text{Im}(z) = -3\).

Equality of Complex Numbers

Two complex numbers are equal if, and only if, their real parts are equal and their imaginary parts are equal.
If \(x_1 + iy_1 = x_2 + iy_2\), then:

\(x_1 = x_2\) and \(y_1 = y_2\)

Key Takeaway: Complex numbers unite real numbers and imaginary numbers into one single system \(x + iy\). Always keep the real and imaginary components separated.

2. Basic Arithmetic Operations

Addition and Subtraction

To add or subtract complex numbers, simply collect the real parts together and the imaginary parts together—just like collecting like terms in algebra.

\((a + ib) + (c + id) = (a + c) + i(b + d)\)
\((a + ib) - (c + id) = (a - c) + i(b - d)\)

Example: \((5 + 2i) + (3 - 7i) = (5 + 3) + (2 - 7)i = 8 - 5i\)

Multiplication

Multiply two complex numbers by expanding brackets just like normal binomials, and replace \(i^2\) with \(-1\):

\((a + ib)(c + id) = ac + iad + ibc + i^2bd\)
Since \(i^2 = -1\), this simplifies to:
\((a + ib)(c + id) = (ac - bd) + i(ad + bc)\)

Example: Calculate \((3 + 2i)(4 - 5i)\)
\(= 12 - 15i + 8i - 10i^2\)
\(= 12 - 7i - 10(-1)\)
\(= 12 - 7i + 10 = 22 - 7i\)

The Complex Conjugate

For any complex number \(z = x + iy\), its complex conjugate is denoted by \(z^*\) (or \(\bar{z}\)) and is found by reversing the sign of the imaginary part:

\(z^* = x - iy\)

When you multiply a complex number by its conjugate, the imaginary terms cancel out completely, yielding a non-negative real number:

\(z z^* = (x + iy)(x - iy) = x^2 - ixy + ixy - i^2 y^2 = x^2 - (-1)y^2 = x^2 + y^2\)

Division (Realising the Denominator)

To divide two complex numbers, multiply both the numerator and the denominator by the complex conjugate of the denominator. This removes all \(i\) terms from the bottom (similar to rationalising a surd denominator).

\(\frac{a + ib}{c + id} = \frac{(a + ib)(c - id)}{(c + id)(c - id)} = \frac{(ac + bd) + i(bc - ad)}{c^2 + d^2}\)

Worked Example: Express \(\frac{7 + 4i}{2 - 3i}\) in the form \(x + iy\).

Step 1: Identify the conjugate of the denominator: \((2 - 3i)^* = 2 + 3i\).
Step 2: Multiply top and bottom by \(2 + 3i\):
\(\frac{7 + 4i}{2 - 3i} = \frac{(7 + 4i)(2 + 3i)}{(2 - 3i)(2 + 3i)}\)
Step 3: Expand the numerator and denominator:
Numerator: \(14 + 21i + 8i + 12i^2 = 14 + 29i - 12 = 2 + 29i\)
Denominator: \(2^2 + (-3)^2 = 4 + 9 = 13\)
Step 4: Separate into real and imaginary parts:
\(= \frac{2}{13} + \frac{29}{13}i\)

Key Takeaway: Always multiply top and bottom by the denominator's conjugate to carry out division. Remember that \(i^2 = -1\), so \((x + iy)(x - iy) = x^2 + y^2\).

3. Argand Diagrams and Modulus-Argument Form

The Argand Diagram

An Argand diagram is a geometric plane where complex numbers are plotted as coordinates or position vectors:

The horizontal axis is the Real Axis (\(\text{Re}\)).
The vertical axis is the Imaginary Axis (\(\text{Im}\)).

A complex number \(z = x + iy\) corresponds to the point \((x, y)\) on the plane or the vector \(\begin{pmatrix} x \\ y \end{pmatrix}\) from the origin \((0, 0)\).

The Modulus of \(z\)

The modulus of \(z\), written as \(|z|\) or \(r\), is the straight-line distance from the origin to the point \((x, y)\). Using Pythagoras' Theorem:

\(r = |z| = \sqrt{x^2 + y^2} \quad (|z| \ge 0)\)

Key properties of the modulus:
\(|z_1 z_2| = |z_1||z_2|\)
\(\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}\)

The Argument of \(z\)

The argument of \(z\), written as \(\arg(z)\) or \(\theta\), is the angle measured from the positive real axis to the line representing \(z\).

Principal Argument Convention: \(-\pi < \arg(z) \le \pi\) radians (or \(-180^\circ < \arg(z) \le 180^\circ\)).
Angles measured anticlockwise are positive (\(0\) to \(\pi\)).
Angles measured clockwise are negative (\(0\) to \(-\pi\)).

How to Find the Argument by Quadrants

First find the basic acute reference angle \(\alpha = \arctan\left(\left|\frac{y}{x}\right|\right)\), then determine \(\theta\) depending on the quadrant:

Quadrant 1 (\(x > 0, y > 0\)): \(\theta = \alpha\)
Quadrant 2 (\(x < 0, y > 0\)): \(\theta = \pi - \alpha\)
Quadrant 3 (\(x < 0, y < 0\)): \(\theta = -(\pi - \alpha) = -\pi + \alpha\)
Quadrant 4 (\(x > 0, y < 0\)): \(\theta = -\alpha\)

Worked Example: Find the modulus and principal argument of \(z = -\sqrt{3} - i\).

Modulus: \(r = \sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2\)
Argument:
Reference angle: \(\alpha = \arctan\left(\left|\frac{-1}{-\sqrt{3}}\right|\right) = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\)
Since both \(x < 0\) and \(y < 0\), the point lies in Quadrant 3.
\(\theta = -\left(\pi - \frac{\pi}{6}\right) = -\frac{5\pi}{6}\)

Modulus-Argument (Polar) Form

Using trigonometry, \(x = r\cos\theta\) and \(y = r\sin\theta\). Substituting these gives:

\(z = r(\cos\theta + i\sin\theta)\)

Multiplication and Division in Modulus-Argument Form

When multiplying or dividing complex numbers in polar form, calculations become remarkably neat:

Multiplication: Multiply moduli, add arguments.
\(|z_1 z_2| = r_1 r_2\) and \(\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)\)
Division: Divide moduli, subtract arguments.
\(\left|\frac{z_1}{z_2}\right| = \frac{r_1}{r_2}\) and \(\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)\)

Note: If adding or subtracting angles takes the result outside the principal range \(-\pi < \theta \le \pi\), add or subtract \(2\pi\) to bring it back into range.

Standard Form Reminder: Ensure the expression is strictly in the form \(r(\cos\theta + i\sin\theta)\). If you have a minus sign in the middle, use the trigonometric symmetries \(\cos(-\theta) = \cos\theta\) and \(\sin(-\theta) = -\sin\theta\) to rewrite:
\(r(\cos\theta - i\sin\theta) = r(\cos(-\theta) + i\sin(-\theta))\)

Key Takeaway: The modulus is the distance from the origin; the argument is the directional angle. Always sketch a small Argand diagram to double-check which quadrant your complex number lives in!

4. Polynomial Equations with Real Coefficients

The Conjugate Root Theorem

For any polynomial equation \(P(z) = 0\) where all coefficients are real numbers, if a complex number \(z = \alpha + i\beta\) (\(\beta \ne 0\)) is a root, then its complex conjugate \(z^* = \alpha - i\beta\) is also a root.

Complex roots of real polynomials always occur in conjugate pairs.

Forming the Quadratic Factor

If \(\alpha + i\beta\) and \(\alpha - i\beta\) are roots, their corresponding real quadratic factor is:

\((z - (\alpha + i\beta))(z - (\alpha - i\beta)) = ((z - \alpha) - i\beta)((z - \alpha) + i\beta)\)
\(= (z - \alpha)^2 - (i\beta)^2 = (z - \alpha)^2 + \beta^2\)
\(= z^2 - 2\alpha z + (\alpha^2 + \beta^2)\)

Solving Quadratics, Cubics, and Quartics

1. Quadratic Equations (\(az^2 + bz + c = 0\))

When the discriminant \(\Delta = b^2 - 4ac < 0\), apply the quadratic formula as normal:

\(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm i\sqrt{4ac - b^2}}{2a}\)

2. Cubic Equations

A cubic polynomial with real coefficients always has 3 roots. It factorises into one real linear factor and one real quadratic factor:

\(P(z) = (z - k)(az^2 + bz + c) = 0\)

This produces one real root and two complex conjugate roots.

3. Quartic Equations

A quartic polynomial with real coefficients has 4 roots. It can be factorised into two real quadratic factors, leading to up to two pairs of complex conjugate roots.

Worked Example: Given that \(z = 2 - 3i\) is a root of the equation \(z^3 - 3z^2 + 9z + 13 = 0\), find all the roots.

Step 1: State the conjugate root.
Since the polynomial has real coefficients, the complex conjugate \(z = 2 + 3i\) must also be a root.

Step 2: Find the quadratic factor.
\((z - (2 - 3i))(z - (2 + 3i)) = (z - 2)^2 + 3^2 = z^2 - 4z + 4 + 9 = z^2 - 4z + 13\)

Step 3: Factorise the cubic.
Divide \(z^3 - 3z^2 + 9z + 13\) by \((z^2 - 4z + 13)\) by inspection or polynomial long division:
\(z^3 - 3z^2 + 9z + 13 = (z + 1)(z^2 - 4z + 13) = 0\)

Step 4: Solve for the remaining root:
\(z + 1 = 0 \implies z = -1\)

Final Roots: \(z = -1, \quad z = 2 - 3i, \quad z = 2 + 3i\)

Key Takeaway: If a polynomial has real coefficients, non-real roots come in conjugate pairs \(\alpha \pm i\beta\). Multiply \((z - (\alpha + i\beta))(z - (\alpha - i\beta))\) to obtain the real quadratic factor \(z^2 - 2\alpha z + (\alpha^2 + \beta^2)\).

5. Quick Summary & Common Pitfalls Checklist

Imaginary part error: Write \(\text{Im}(z) = y\), not \(yi\).
Sign error in division: Remember that \((c + id)(c - id) = c^2 + d^2\) with a plus sign in the middle.
Quadrant check: Don't just press \(\arctan\) on the calculator for arguments. Sketch the point on an Argand diagram to adjust for Quadrants 2, 3, or 4.
Conjugate root condition: Conjugate roots only occur in pairs if all polynomial coefficients are real numbers.
Principal argument limits: Ensure final arguments satisfy \(-\pi < \theta \le \pi\).

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the definition of the imaginary unit \(i\) and how is it used to simplify square roots of negative numbers?

The imaginary unit is defined by \(i = \sqrt{-1}\), which means \(i^2 = -1\). Square roots of negative numbers simplify as \(\sqrt{-k} = \sqrt{k}i\) for \(k > 0\).

What is the Cartesian form of a complex number, and what are its real and imaginary parts?

The Cartesian form is \(z = x + iy\), where \(x, y \in \mathbb{R}\). The real part is \(\text{Re}(z) = x\) and the imaginary part is \(\text{Im}(z) = y\) (not \(yi\)).

What condition must be met for two complex numbers to be equal?

Two complex numbers are equal if, and only if, their real parts are equal and their imaginary parts are equal: \(x_1 + iy_1 = x_2 + iy_2 \iff x_1 = x_2\) and \(y_1 = y_2\).

How do you add, subtract, and multiply complex numbers in Cartesian form?

Add or subtract by collecting real and imaginary terms separately. Multiply by expanding brackets algebraically and replacing \(i^2\) with \(-1\).

What is the complex conjugate of \(z = x + iy\), and what is the result of multiplying \(z\) by its conjugate?

The complex conjugate is \(z^* = x - iy\). Multiplying a number by its conjugate gives a non-negative real value: \(z z^* = x^2 + y^2\).

How do you divide one complex number by another in Cartesian form?

Multiply both the numerator and denominator by the complex conjugate of the denominator to realise the denominator, then simplify.

What is an Argand diagram?

An Argand diagram is a geometric coordinate plane where the horizontal axis represents the real part (\(\text{Re}\)) and the vertical axis represents the imaginary part (\(\text{Im}\)).

What is the modulus of a complex number \(z = x + iy\)?

The modulus \(|z|\) (or \(r\)) is the distance from the origin to \(z\) on the Argand plane: \(|z| = \sqrt{x^2 + y^2}\).

What is the principal argument of a complex number and what is its standard range?

The principal argument \(\arg(z) = \theta\) is the angle measured from the positive real axis to the line representing \(z\), defined in the range \(-\pi < \theta \le \pi\) radians.

What is the modulus-argument form of a complex number and how do you multiply and divide numbers in this form?

Modulus-argument form is \(z = r(\cos\theta + i\sin\theta)\). When multiplying, multiply moduli and add arguments; when dividing, divide moduli and subtract arguments.

What is the Conjugate Root Theorem for polynomial equations?

For any polynomial equation with real coefficients, if \(z = \alpha + i\beta\) is a root, its complex conjugate \(z^* = \alpha - i\beta\) must also be a root.

How do you form a real quadratic factor from a pair of complex conjugate roots \(\alpha \pm i\beta\)?

The corresponding real quadratic factor is \((z - (\alpha + i\beta))(z - (\alpha - i\beta)) = z^2 - 2\alpha z + (\alpha^2 + \beta^2)\).

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