# Complex Numbers

> CIE A-Level Mathematics · Pure Mathematics 3
> Source: https://www.owlsprep.com/study/cie-9709-u2-complex-numbers/

This subtopic extends the real number system to introduce complex numbers, enabling solutions to equations with no real roots. You will learn basic arithmetic, key properties (modulus, argument, conjugate) and representation on the Argand diagram.

**Prerequisites:** [Quadratic equations and discriminants](https://www.owlsprep.com/study/cie-9709-u1-quadratic-equations/); [Coordinate geometry basics](https://www.owlsprep.com/study/cie-9709-u1-coordinate-geometry/)

## Learning objectives

- Recall the definition of complex numbers and the imaginary unit $i$
- Perform arithmetic operations on complex numbers in Cartesian form
- Calculate the modulus, argument and conjugate of any complex number
- Apply the conjugate root theorem to polynomials with real coefficients
- Find the square roots of a complex number using simultaneous equations
- Express complex numbers in modulus-argument (polar) and exponential form, and multiply and divide them in that form
- Interpret conjugation, addition and multiplication geometrically on the Argand diagram
- Sketch and describe loci defined by $|z-a|=r$, $|z-a|=|z-b|$ and $\arg(z-a)=\alpha$, including simple inequality regions

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 P3 past papers (2016–2025): **Complex Numbers** appears **82 times** in the last 10 years — **12.6%** of all P3 questions (82 of 651).

Most-tested forms: Complex numbers: Cartesian form, conjugate, arithmetic (50), Loci of complex numbers on the Argand diagram (46), Modulus, argument, polar form (35), Roots of polynomials with real coefficients (conjugate pairs) (12).

## Definition and Basic Representation

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

*Notation:* z = x + iy

*Example:* For $z = 3 - 2i$, $
\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 $z_1 = 2 + 3i$ and $z_2 = -1 - 4i$ on an Argand diagram, and state $
\text{Re}(z_1)$ and $
\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 $z_2 = -1 - 4i$, 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.

## Arithmetic of Complex Numbers

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 $i^2 = -1$.

**Complex Conjugate** — For $z = x + iy$, the conjugate is $z^* = x - iy$, found by flipping the sign of the imaginary part. A key property is $zz^* = |z|^2$, a real number.

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

*Example:* Conjugate of $5 + 2i$ is $5 - 2i$.

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

1. The conjugate of the denominator $1 - 4i$ is $1 + 4i$. 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. $$\frac{-5 + 14i}{17} = -\frac{5}{17} + \frac{14}{17}i$$

> **warning**
>
> Always remember to flip the sign of the imaginary part correctly when finding the conjugate. A common error is flipping the sign of the real part instead.

## Square Roots of a Complex Number

Every non-zero complex number has exactly **two** square roots. To find $\sqrt{a+ib}$ in Cartesian form, set $(x+iy)^2 = a + ib$, expand, and compare real and imaginary parts to get two simultaneous equations.

**Square Roots by Comparing Parts** — Expanding gives $(x^2 - y^2) + 2xyi = a + ib$. Comparing parts: $x^2 - y^2 = a$ and $2xy = b$. Solve this pair for **real** $x, y$; the two solutions are negatives of each other.

*Notation:* (x+iy)^2 = a + ib

*Example:* For $\sqrt{5+12i}$: $x^2 - y^2 = 5$ and $2xy = 12$.

**Worked example:** Find the square roots of $5 + 12i$, giving each in the form $x + iy$.

1. Let $\sqrt{5+12i} = x + iy$ with $x, y$ real. Square both sides:
2. $$(x+iy)^2 = x^2 - y^2 + 2xyi = 5 + 12i$$
3. Compare real and imaginary parts to form two equations:
4. $$x^2 - y^2 = 5, \qquad 2xy = 12$$
5. From the second equation, $y = \frac{6}{x}$. Substitute into the first:
6. $$x^2 - \frac{36}{x^2} = 5 \implies x^4 - 5x^2 - 36 = 0$$
7. This is a quadratic in $x^2$: $(x^2 - 9)(x^2 + 4) = 0$. Since $x$ is real, $x^2 = 9$, so $x = \pm 3$.
8. For $x = 3$: $y = \frac{6}{3} = 2$. For $x = -3$: $y = \frac{6}{-3} = -2$.
9. The two square roots are $3 + 2i$ and $-3 - 2i$, i.e. $\pm(3 + 2i)$.

> **tip**
>
> Check by squaring: $(3+2i)^2 = 9 + 12i + 4i^2 = 5 + 12i$. The two roots always differ only by an overall sign, so once you have one, the other is its negative.

> **Exam tip:** Keep $x$ and $y$ real: reject any solution where $x^2$ comes out negative. Using $2xy = b$ to fix the sign of $y$ relative to $x$ avoids spurious answers.

## Modulus and Principal Argument

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 $-\pi < \arg(z) \leq \pi$ for CIE exams.

**Modulus and Argument** — For $z = x + iy$: $|z| = \sqrt{x^2 + y^2}$, and $\tan(\arg(z)) = \frac{y}{x}$. Always adjust the angle to match the quadrant of the complex number.

*Notation:* $|z|$ (modulus), $\arg(z)$ (argument)

**Worked example:** Find the modulus and principal argument of $z = -1 + i\sqrt{3}$. Give your answer in radians.

1. Identify $x = -1$ (real part) and $y = \sqrt{3}$ (imaginary part).
2. Calculate the modulus:
3. $$|z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$$
4. Find the reference angle: $\tan\theta = \frac{y}{x} = \frac{\sqrt{3}}{-1} = -\sqrt{3}$, so reference angle is $\frac{\pi}{3}$.
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: $|z| = 2$, $\arg(z) = \frac{2\pi}{3}$, which is within the range for principal argument.

> **tip**
>
> A quick check: if your argument is outside $(-\pi, \pi]$, add or subtract $2\pi$ to get the principal value.

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

## Modulus-Argument (Polar) Form

Instead of $x + iy$, a complex number can be written using its modulus $r = |z|$ and argument $\theta = \arg(z)$. This **modulus-argument (polar) form** makes multiplication and division especially simple.

**Modulus-Argument (Polar) and Exponential Form** — Here $r = |z| = \sqrt{x^2 + y^2} \geq 0$ and $\theta = \arg(z)$ taken in the principal range $-\pi < \theta \leq \pi$. The exponential form $re^{i\theta}$ is an equivalent notation for the same number, with $x = r\cos\theta$ and $y = r\sin\theta$.

*Notation:* z = r(\cos\theta + i\sin\theta) = re^{i\theta}

*Example:* $-1 + i\sqrt{3} = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) = 2e^{i\,2\pi/3}$.

In polar form, **multiplying multiplies the moduli and adds the arguments**, while **dividing divides the moduli and subtracts the arguments**. This is far quicker than expanding brackets.

**Multiplication and Division in Polar Form** — $|z_1 z_2| = |z_1||z_2|$ and $\arg(z_1 z_2) = \arg z_1 + \arg z_2$; $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$ and $\arg\!\left(\frac{z_1}{z_2}\right) = \arg z_1 - \arg z_2$. After adding or subtracting, adjust the result by $\pm 2\pi$ so it lies in $-\pi < \theta \leq \pi$.

*Example:* If $\arg z_1 = \frac{3\pi}{4}$ and $\arg z_2 = \frac{3\pi}{4}$, then $\arg(z_1 z_2) = \frac{3\pi}{2}$, adjusted to $\frac{3\pi}{2} - 2\pi = -\frac{\pi}{2}$.

**Worked example:** Let $z_1 = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)$ and $z_2 = 3\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$. Find $z_1 z_2$ and $\frac{z_1}{z_2}$ in modulus-argument form.

1. For the product, multiply the moduli and add the arguments:
2. $$|z_1 z_2| = 2 \times 3 = 6, \qquad \arg(z_1 z_2) = \frac{\pi}{3} + \frac{\pi}{4} = \frac{7\pi}{12}$$
3. $$z_1 z_2 = 6\left(\cos\frac{7\pi}{12} + i\sin\frac{7\pi}{12}\right)$$
4. For the quotient, divide the moduli and subtract the arguments:
5. $$\left|\frac{z_1}{z_2}\right| = \frac{2}{3}, \qquad \arg\!\left(\frac{z_1}{z_2}\right) = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}$$
6. $$\frac{z_1}{z_2} = \frac{2}{3}\left(\cos\frac{\pi}{12} + i\sin\frac{\pi}{12}\right)$$
7. Both arguments $\frac{7\pi}{12}$ and $\frac{\pi}{12}$ already lie in $-\pi < \theta \leq \pi$, so no adjustment is needed.

> **tip**
>
> The exponential form makes these rules obvious: $r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} = r_1 r_2\, e^{i(\theta_1 + \theta_2)}$ and $\frac{r_1 e^{i\theta_1}}{r_2 e^{i\theta_2}} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)}$, using ordinary index laws.

> **Exam tip:** After adding or subtracting arguments, always bring the result back into the principal range $-\pi < \theta \leq \pi$ before stating the final argument.

## Geometric Effects on the Argand Diagram

Each algebraic operation on complex numbers has a clear geometric meaning on the Argand diagram, which helps you picture and check your answers.

**Geometric Interpretations** — **Conjugation** $z \mapsto z^*$ reflects the point in the real axis. **Addition** $z_1 + z_2$ is vector addition (the parallelogram rule). **Multiplication** by $z_2$ scales the modulus by $|z_2|$ and rotates by $\arg z_2$ about the origin.

*Example:* Multiplying by $i$ (modulus 1, argument $\frac{\pi}{2}$) rotates a point $90^\circ$ anticlockwise about the origin without changing its distance from the origin.

**Worked example:** The point $A$ represents $z = 3 + i$. Describe geometrically the points representing (i) $z^*$, (ii) $iz$, and give each in Cartesian form.

1. (i) $z^* = 3 - i$. Geometrically this is the reflection of $A$ in the real axis: the real part is unchanged and the imaginary part changes sign.
2. (ii) Multiplying by $i$ rotates $A$ by $\frac{\pi}{2}$ (90°) anticlockwise about the origin and keeps $|z|$ the same, since $|i| = 1$.
3. $$iz = i(3 + i) = 3i + i^2 = -1 + 3i$$
4. Check the rotation: $z = 3 + i$ maps to $(3,1) \to (-1, 3)$, which is indeed a $90^\circ$ anticlockwise turn about the origin.

> **tip**
>
> Because multiplication adds arguments, repeatedly multiplying by a unit-modulus number just keeps rotating the point around a circle centred at the origin.

## Loci in the Argand Diagram

A **locus** is the set of all points $z$ satisfying a given condition. Three standard conditions appear repeatedly in Paper 3, each with a simple geometric shape. Write $z = x + iy$ and interpret $|z - a|$ as the distance from $z$ to the fixed point $a$.

**Three Standard Loci** — **Circle:** $|z - a| = r$ is a circle with centre $a$ and radius $r$. **Perpendicular bisector:** $|z - a| = |z - b|$ is the perpendicular bisector of the line segment joining $a$ and $b$ (all points equidistant from $a$ and $b$). **Half-line:** $\arg(z - a) = \alpha$ is a ray (half-line) starting at $a$ — with the endpoint $a$ excluded — making angle $\alpha$ with the positive real direction.

*Example:* $|z - (1 + 2i)| = 3$ is a circle of radius 3 centred at the point $(1, 2)$.

Replacing $=$ with an inequality gives a **region**. For example, $|z - a| < r$ is the open disc *inside* the circle (boundary not included), while $|z - a| \leq r$ includes the boundary circle.

**Worked example:** Describe and sketch the locus of $z$ given by (a) $|z - (2 + i)| = 4$, (b) $|z - 3| = |z + i|$, (c) $\arg(z - 1) = \frac{\pi}{4}$.

1. (a) This is the set of points a distance 4 from $2 + i$. It is a **circle** with centre $(2, 1)$ and radius 4.
2. (b) Write it as $|z - 3| = |z - (-i)|$: points equidistant from $a = 3$ (the point $(3, 0)$) and $b = -i$ (the point $(0, -1)$). It is the **perpendicular bisector** of the segment joining $(3, 0)$ and $(0, -1)$.
3. (c) Points where $z - 1$ has argument $\frac{\pi}{4}$. This is a **half-line** starting at the point $(1, 0)$ (open endpoint) going up and to the right at $45^\circ$ to the positive real axis.
4. On an Argand diagram: draw the circle for (a), the straight bisector line for (b), and the ray from $(1,0)$ for (c) with an open circle marking the excluded start point.

> **warning**
>
> For a half-line $\arg(z - a) = \alpha$, the endpoint $a$ itself is **not** included (the argument of $0$ is undefined), so mark it with an open circle. Only the single ray at angle $\alpha$ counts — not the opposite direction $\alpha + \pi$.

> **Exam tip:** Always convert $|z - a|$ into 'distance from the point $a$'. For $|z + i|$, rewrite as $|z - (-i)|$ first so you read off the correct centre or point.

## Complex Conjugate Root Theorem

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.

**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 $2 - 3i$. Find the quadratic equation.

1. By the conjugate root theorem, the second root is the conjugate of $2 - 3i$, which is $2 + 3i$.
2. Calculate the sum of the roots: $(2 - 3i) + (2 + 3i) = 4$.
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 $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$, substitute the values:
6. $$x^2 - 4x + 13 = 0$$

## Common pitfalls

- **Wrong:** Claiming $i^2 = 1$ by using $i = \sqrt{-1}$ so $i^2 = \sqrt{(-1)(-1)} = \sqrt{1} = 1$
  - Why it fails: The rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ only applies for non-negative real $a$ and $b$, so this extension is invalid.
  - Correct: Always use the definition $i^2 = -1$ directly, do not extend real-number root rules to negative numbers.
- **Wrong:** Writing $\arg(z) = \arctan(\frac{y}{x})$ directly without adjusting for quadrant
  - Why it fails: $\arctan$ only returns values between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, so it will give the wrong angle for complex numbers in the second and third quadrants.
  - Correct: Always check the quadrant of your complex number, and add or subtract $\pi$ as needed to get the principal argument.
- **Wrong:** When dividing complex numbers, multiply only the numerator by the conjugate of the denominator
  - Why it fails: This changes the value of the fraction, leading to an incorrect result.
  - Correct: Always multiply both the numerator and denominator by the conjugate to keep the fraction equivalent.
- **Wrong:** Using the conjugate root theorem for polynomials with complex coefficients
  - Why it fails: The theorem only holds when all coefficients of the polynomial are real.
  - Correct: Only use the conjugate root theorem if the question explicitly states the polynomial has real coefficients.
- **Wrong:** Giving only one square root of a complex number, or keeping a solution where $x^2$ comes out negative.
  - Why it fails: Every non-zero complex number has exactly two square roots, and $x, y$ must be real, so a negative $x^2$ has no real solution.
  - Correct: Solve $x^2 - y^2 = a$ and $2xy = b$ for real $x, y$, reject non-real solutions, and give both roots $\pm(x + iy)$.
- **Wrong:** Leaving $\arg(z_1 z_2)$ or $\arg(z_1/z_2)$ outside $-\pi < \theta \leq \pi$ after adding or subtracting arguments.
  - Why it fails: The sum or difference of two principal arguments can fall outside the principal range, so it is no longer the principal argument.
  - Correct: After adding or subtracting, add or subtract $2\pi$ as needed to bring the argument back into $-\pi < \theta \leq \pi$.
- **Wrong:** Sketching $\arg(z - a) = \alpha$ as a full straight line through $a$, or including the point $a$ itself.
  - Why it fails: The argument fixes a single direction from $a$, so only one ray qualifies; and $\arg 0$ is undefined, so $a$ is excluded.
  - Correct: Draw a single half-line from $a$ at angle $\alpha$, mark the endpoint $a$ with an open circle, and do not draw the opposite direction $\alpha + \pi$.

## Cheatsheet

| Property | Rule for $z = x + iy, \; z_1 = x_1 + iy_1, \; z_2 = x_2 + iy_2$ |
| --- | --- |
| 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 | $z_1z_2 = (x_1x_2 - y_1y_2) + i(x_1y_2 + x_2y_1)$ |
| Division | $\frac{z_1}{z_2} = \frac{z_1z_2^*}{\|z_2\|^2}$ |
| Conjugate | $z^* = x - iy, \; zz^* = \|z\|^2$ (always real) |
| Modulus | $\|z\| = \sqrt{x^2 + y^2}$ |
| Argument | $\tan\arg(z) = \frac{y}{x}$, adjust for quadrant, $-\pi < \arg(z) \leq \pi$ |
| Polar form | $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$, $r = \|z\|$, $\theta = \arg(z)$ |
| Polar multiply | $\|z_1z_2\| = \|z_1\|\|z_2\|$, $\arg(z_1z_2) = \arg z_1 + \arg z_2$ |
| Polar divide | $\left\|\frac{z_1}{z_2}\right\| = \frac{\|z_1\|}{\|z_2\|}$, $\arg\!\left(\frac{z_1}{z_2}\right) = \arg z_1 - \arg z_2$ |
| Square roots | $\sqrt{a+ib} = x+iy$: solve $x^2 - y^2 = a$, $2xy = b$; roots $\pm(x+iy)$ |
| Locus: circle | $\|z - a\| = r$: circle, centre $a$, radius $r$ |
| Locus: bisector | $\|z - a\| = \|z - b\|$: perpendicular bisector of $a$, $b$ |
| Locus: half-line | $\arg(z - a) = \alpha$: half-line from $a$ (excl.), angle $\alpha$ |
| Conjugate Root Theorem | Complex roots come in conjugate pairs for real polynomials |

## What's next

This subtopic now covers the full range of complex-number skills tested in CIE 9709 Paper 3: Cartesian arithmetic, the modulus, argument and conjugate, square roots, modulus-argument (polar) and exponential form with multiplication and division, the geometric effects of these operations, and loci on the Argand diagram. (De Moivre's theorem and the nth roots of a complex number are NOT part of 9709 — they belong to Further Mathematics 9231.) Complex numbers regularly appear in combined questions with polynomials and trigonometry, so a solid command of every part of this subtopic is critical for a high grade in Paper 3.

- [Mechanics](https://www.owlsprep.com/study/cie-9709-u3-overview/)
- [Kinematics](https://www.owlsprep.com/study/cie-9709-u3-kinematics/)
- [Forces and Equilibrium](https://www.owlsprep.com/study/cie-9709-u3-forces-and-equilibrium/)

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