Complex Numbers

CIE A-Level Mathematics· 20 min read

1. Definition and Basic Representation★★☆☆☆⏱ 5 min

📘 Definition

Complex Number

z = x + iy

A number of the form $x + i y$ where $x, y \in R$ , and $i^{2} = - 1$ . $x$ is the real part, $y$ is the imaginary part.

Example:

For $z = 3 - 2 i$ , $ \text{Re}(z) = 3 $,$ \text{Im}(z) = -2$.

A complex number can be plotted as a point $(x, y)$ on the Argand diagram: a coordinate plane with a horizontal real axis and vertical imaginary axis.

📐 Worked Example

Plot $z_{1} = 2 + 3 i$ and $z_{2} = - 1 - 4 i$ on an Argand diagram, and state $ \text{Re}(z_1) $an d$ \text{Im}(z_2)$.

1
Extract coordinates from the Cartesian form: $z_{1}$ has real part 2, imaginary part 3, so it maps to the point $(2, 3)$ .
2
For $z_{2} = - 1 - 4 i$ , the real part is $- 1$ and imaginary part is $- 4$ , so it maps to the point $(- 1, - 4)$ .
3
Plot both points on the grid, with the x-axis labelled 'Real' and y-axis labelled 'Imaginary'.
4
Final answer: $ \text{Re}(z_1) = 2 $,$ \text{Im}(z_2) = -4$.

Exam tip:

Always clearly label axes and points when drawing an Argand diagram to earn full method marks.

2. Arithmetic of Complex Numbers★★★☆☆⏱ 7 min

Addition and subtraction of complex numbers follow simple rules: add/subtract the real parts and imaginary parts separately. For multiplication, expand like any binomial and simplify using $i^{2} = - 1$ .

📘 Definition

Complex Conjugate

$z^{*}$ or $\overline{z}$

For $z = x + i y$ , the conjugate is $z^{*} = x - i y$ , found by flipping the sign of the imaginary part. A key property is $z z^{*} = ∣ z ∣^{2}$ , a real number.

Example:

Conjugate of $5 + 2 i$ is $5 - 2 i$ .

To divide two complex numbers, multiply the numerator and denominator by the conjugate of the denominator to turn the denominator into a real number.

📐 Worked Example

Simplify $\frac{3 + 2 i}{1 - 4 i}$ into Cartesian form $x + i y$ .

1
The conjugate of the denominator $1 - 4 i$ is $1 + 4 i$ . Multiply numerator and denominator by this conjugate:
2
$\frac{(3 + 2i)(1 + 4i)}{(1 - 4i)(1 + 4i)}$
3
Expand the denominator using the difference of squares: $(a - b) (a + b) = a^{2} - b^{2}$ :
4
$(1)^2 - (4i)^2 = 1 - 16i^2 = 1 - 16(-1) = 17$
5
Expand the numerator:
6
$3(1) + 3(4i) + 2i(1) + 2i(4i) = 3 + 12i + 2i + 8i^2 = 3 + 14i - 8 = -5 + 14i$
7
Simplify to get Cartesian form:
8
$\frac{-5 + 14i}{17} = -\frac{5}{17} + \frac{14}{17}i$

3. Modulus and Principal Argument★★★★☆⏱ 6 min

Any non-zero complex number can be described by its modulus (distance from the origin) and argument (angle from the positive real axis). The principal argument is defined to lie in the range $- π < ar g (z) \leq π$ for CIE exams.

📘 Definition

Modulus and Argument

$∣ z ∣$ (modulus), $ar g (z)$ (argument)

For $z = x + i y$ : $∣ z ∣ = x^{2} + y^{2}$ , and $tan (ar g (z)) = \frac{y}{x}$ . Always adjust the angle to match the quadrant of the complex number.

📐 Worked Example

Find the modulus and principal argument of $z = - 1 + i 3$ . Give your answer in radians.

1
Identify $x = - 1$ (real part) and $y = 3$ (imaginary part).
2
Calculate the modulus:
3
$|z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$
4
Find the reference angle: $tan θ = \frac{y}{x} = \frac{3}{- 1} = - 3$ , so reference angle is $\frac{π}{3}$ .
5
Check the quadrant: $x < 0$ , $y > 0$ so $z$ lies in the second quadrant.
6
$\arg(z) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
7
Final answer: $∣ z ∣ = 2$ , $ar g (z) = \frac{2 π}{3}$ , which is within the range for principal argument.

Exam tip:

Never just write the calculator output of $arctan (\frac{y}{x})$ as your argument. Always adjust for the correct quadrant.

4. Complex Conjugate Root Theorem★★★☆☆⏱ 5 min

When a polynomial has all real coefficients, any complex root must have its conjugate as another root. This property lets us find unknown roots and construct polynomials when only one complex root is known.

📘 Definition

Complex Conjugate Root Theorem

If $p (x)$ is a polynomial with real coefficients, and $z = a + ib$ is a root of $p (x) = 0$ , then its conjugate $z^{*} = a - ib$ is also a root.

📐 Worked Example

A quadratic equation with real coefficients has one root $2 - 3 i$ . Find the quadratic equation.

1
By the conjugate root theorem, the second root is the conjugate of $2 - 3 i$ , which is $2 + 3 i$ .
2
Calculate the sum of the roots: $(2 - 3 i) + (2 + 3 i) = 4$ .
3
Calculate the product of the roots, using difference of squares:
4
$(2 - 3i)(2 + 3i) = 2^2 - (3i)^2 = 4 - (-9) = 13$
5
For a quadratic $x^{2} - (sum of roots) x + (product of roots) = 0$ , substitute the values:
6
$x^2 - 4x + 13 = 0$

5. Common Pitfalls

Wrong move:

Claiming $i^{2} = 1$ by using $i = - 1$ so $i^{2} = (- 1) (- 1) = 1 = 1$

Why:

The rule $a b = ab$ only applies for non-negative real $a$ and $b$ , so this extension is invalid.

Correct move:

Always use the definition $i^{2} = - 1$ directly, do not extend real-number root rules to negative numbers.

Wrong move:

Writing $ar g (z) = arctan (\frac{y}{x})$ directly without adjusting for quadrant

Why:

$arctan$ only returns values between $- \frac{π}{2}$ and $\frac{π}{2}$ , so it will give the wrong angle for complex numbers in the second and third quadrants.

Correct move:

Always check the quadrant of your complex number, and add or subtract $π$ as needed to get the principal argument.

Wrong move:

When dividing complex numbers, multiply only the numerator by the conjugate of the denominator

Why:

This changes the value of the fraction, leading to an incorrect result.

Correct move:

Always multiply both the numerator and denominator by the conjugate to keep the fraction equivalent.

Wrong move:

Using the conjugate root theorem for polynomials with complex coefficients

Why:

The theorem only holds when all coefficients of the polynomial are real.

Correct move:

Only use the conjugate root theorem if the question explicitly states the polynomial has real coefficients.

6. Quick Reference Cheatsheet

Property	Rule for $z = x + i y, z_{1} = x_{1} + i y_{1}, z_{2} = x_{2} + i y_{2}$
Addition	$z_{1} + z_{2} = (x_{1} + x_{2}) + i (y_{1} + y_{2})$
Subtraction	$z_{1} - z_{2} = (x_{1} - x_{2}) + i (y_{1} - y_{2})$
Multiplication	$z_{1} z_{2} = (x_{1} x_{2} - y_{1} y_{2}) + i (x_{1} y_{2} + x_{2} y_{1})$
Division	$\frac{z _{1}}{z _{2}} = \frac{z _{1} z _{2}^{*}}{∣ z _{2} ∣ ^{2}}$
Conjugate	$z^{} = x - i y, z z^{} = ∣ z ∣^{2}$ (always real)
Modulus	$∣ z ∣ = x^{2} + y^{2}$
Argument	$tan ar g (z) = \frac{y}{x}$ , adjust for quadrant, $- π < ar g (z) \leq π$
Conjugate Root Theorem	Complex roots come in conjugate pairs for real polynomials

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 3
Modulus-argument conversion
2022 · 3
Find quadratic from complex root
2021 · 3
Complex number division

What's Next

This subtopic covers the foundational concepts of complex numbers required for all further work on this topic in CIE Paper 3. Mastering these basics will make learning more advanced concepts like polar form, loci on the Argand diagram, and de Moivre's theorem much more straightforward. Complex numbers regularly appear in combined questions with polynomials and trigonometry, so a solid understanding of this subtopic is critical for achieving a high grade in Paper 3.