$i$ is the imaginary unit, defined as $i = \sqrt{-1}$. It's not a 'real' number because no real number can be squared to produce a negative result. It's a foundational concept that extends our number system.

Why are there so many forms for complex numbers (Cartesian, polar, exponential)?

Each form is useful for different operations. Cartesian form ($x+iy$) is best for addition and subtraction. Polar ($r(\cos\theta + i\sin\theta)$) and exponential ($re^{i\theta}$) forms are far superior for multiplication, division, and finding powers and roots.

My calculator has a complex number mode. Can I just use that in the exam?

You can use it to check your answers, but you cannot rely on it for your working. Exam questions are designed to test your understanding of the methods, such as finding the modulus and argument or applying De Moivre's theorem. You must show these steps clearly to earn method marks.

How do I divide complex numbers in Cartesian form?

To divide $\frac{z_1}{z_2}$, you multiply the numerator and denominator by the complex conjugate of the denominator, $z_2^*$. For example, to compute $\frac{2+3i}{1-i}$, you would calculate $\frac{(2+3i)(1+i)}{(1-i)(1+i)}$. This process is called 'realising the denominator' because the new denominator $(1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 2$ will be a real number.

Where can I learn Complex numbers for Cambridge 9709?

This free lesson on MarkScheme covers syllabus point 3.9 (Complex numbers) for Cambridge A-Level Mathematics. It includes visual revision, key points, and links to real past-paper questions with mark-scheme marking at markscheme.app/courses/9709/3-9-complex-numbers.

Is Complex numbers (3.9) on 9709 free to study?

Yes. The Mathematics 9709 course on MarkScheme is 100% free — including premium visual lessons, exam tips, and past-paper marking. No paywall for Complex numbers.

Complex numbers 9709 — Free A-Level Maths

Complex numbers

Welcome to the world of complex numbers. For centuries, mathematicians were stumped by the simple equation $x^2 = -1$ . There is no 'real' number that squares to give a negative. The solution was to invent a new one: the imaginary unit, $i = \sqrt{-1}$ . This seemingly abstract idea unlocks a powerful branch of mathematics with applications in engineering, physics, and signal processing. In this topic, we'll learn how to define, visualise, and manipulate these fascinating numbers.

BY THE END, YOU CAN…

1Understand and use the Cartesian form (x + iy) and polar form (r(cos θ + i sin θ) = re^(iθ)) of a complex number.
2Represent complex numbers on an Argand diagram and calculate their modulus and argument.
3Perform operations (multiplication, division, powers) on complex numbers in different forms.
4Apply De Moivre's theorem to find powers of complex numbers and solve related problems.

In simple terms

A friendly intro before the formal notes — no formulas yet.

Numbers in 2D

Complex numbers extend our number system into two dimensions, allowing us to solve equations that have no real solutions. They can be visualised as points on a plane, with a distance from the origin (modulus) and an angle (argument).

THINK OF IT LIKE…

Imagine giving directions. Instead of saying 'go 3 miles East, then 4 miles North', you could say 'go 5 miles at a bearing of 53 degrees'. The first is like the Cartesian form (3 + 4i), giving horizontal and vertical components. The second is like the polar form, giving a direct distance and angle. Both describe the same final location.

1
A complex number z = x + iy is plotted as the point (x, y) in the Argand diagram, with a real (x) and imaginary (y) axis.
2
The modulus |z| is the distance from the origin, calculated as √(x² + y²). The argument, arg z, is the angle from the positive real axis.
3
The polar form uses the modulus (r) and argument (θ) to write z = r(cos θ + i sin θ), which simplifies to the exponential form z = re^(iθ).
4
De Moivre’s theorem provides a powerful shortcut for powers: (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. This means to raise a complex number to a power, you raise the modulus to that power and multiply the argument.

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

z = x + iy

Tap a symbol — great for exam definitions

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

Tap a symbol — great for exam definitions

$[r(\cos\theta + i\sin\theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta))$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

Cartesian Form and the Argand Diagram

A complex number, $z$ , is written in Cartesian form as $z = x + iy$ . Here, $x$ is the 'real part' and $y$ is the 'imaginary part'. We can visualise this number not on a number line, but on a 2D plane called an Argand diagram. The horizontal axis represents the real part ( $x$ ) and the vertical axis represents the imaginary part ( $y$ ). So, the complex number $z = 3 + 2i$ is plotted at the coordinate $(3, 2)$ .

Modulus and Argument

Since a complex number is a point on a plane, we can describe its position in two main ways. The first is with Cartesian coordinates $(x, y)$ . The second is with polar coordinates, which involve distance and direction. The distance from the origin $(0,0)$ to the point $(x,y)$ is called the modulus, denoted $|z|$ or $r$ . The angle made with the positive real axis, measured anticlockwise, is the argument, denoted $\arg(z)$ or $\theta$ .

For $z = x + iy$ : Modulus: $|z| = r = \sqrt{x^2 + y^2}$ Argument: $\tan(\theta) = \frac{y}{x}$

The modulus is always non-negative, as it represents a distance.
The argument must be calculated carefully. Always sketch the point on an Argand diagram to ensure you find the angle in the correct quadrant.
The principal argument is the unique value of $\theta$ such that $-\pi < \theta \le \pi$ . This is the standard range required in Cambridge exams unless specified otherwise.

Polar and Exponential Forms

Using the modulus $r$ and argument $\theta$ , we can express any complex number in polar form. From trigonometry on the Argand diagram, we can see that $x = r\cos\theta$ and $y = r\sin\theta$ . Substituting these into the Cartesian form gives $z = r\cos\theta + i(r\sin\theta) = r(\cos\theta + i\sin\theta)$ . A more compact and extremely useful notation is the exponential form, derived from Euler's formula ( $e^{i\theta} = \cos\theta + i\sin\theta$ ). These forms make multiplication and division much simpler.

Polar Form: $z = r(\cos\theta + i\sin\theta)$ Exponential Form: $z = re^{i\theta}$

Rules for Operations: If $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$ : $z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}$ $\frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)}$

EXAM TIP

When a question asks for the argument, it implies the principal argument unless stated otherwise. After performing multiplication or division, your resulting argument might be outside the range $(-\pi, \pi]$ . You must add or subtract multiples of $2\pi$ to bring it back into the principal range.

De Moivre's Theorem

De Moivre's theorem provides an elegant way to find powers of complex numbers. It follows directly from the multiplication rule for exponential form. If we want to find $z^n$ where $z = re^{i\theta}$ , we get $(re^{i\theta})^n = r^n (e^{i\theta})^n = r^n e^{in\theta}$ . This tells us to raise the modulus to the power $n$ and multiply the argument by $n$ .

De Moivre's Theorem: $[r(\cos\theta + i\sin\theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta))$

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

Let the complex number $u = -\sqrt{3} + i$ . (i) Find the modulus and argument of $u$ . (ii) Express $u$ in exponential form.

1
Given $u = -\sqrt{3} + i$ . This is in the form $x+iy$ with $x = -\sqrt{3}$ and $y = 1$ . (i) Modulus and Argument First, calculate the modulus: $|u| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2$ . [M1, A1] Next, find the argument $\theta$ . Let's first find the related acute angle $\alpha$ : $\alpha = \tan^{-1} \left| \frac{y}{x} \right| = \tan^{-1} \left| \frac{1}{-\sqrt{3}} \right| = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6}$ . [M1] Now, sketch an Argand diagram. The point $(-\sqrt{3}, 1)$ is in the second quadrant. For the second quadrant, the principal argument is $\theta = \pi - \alpha$ . $\arg(u) = \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ . [A1] (ii) Exponential Form The exponential form is $z = re^{i\theta}$ . We have $r=2$ and $\theta = \frac{5\pi}{6}$ . So, $u = 2e^{i(5\pi/6)}$ . [B1 ft]

✓ FULLY WORKED

EXAMPLE 2

Use De Moivre's theorem to find $(1 - i)^8$ .

1
To use De Moivre's theorem, we must first convert $z = 1 - i$ into polar or exponential form. Step 1: Find modulus and argument of $z=1-i$ . $x=1, y=-1$ . Modulus: $|z| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$ . [M1] Argument: The point $(1, -1)$ is in the fourth quadrant. $\theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4}$ . [M1] So, $z = \sqrt{2} \left( \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) \right)$ . Step 2: Apply De Moivre's Theorem. We want to find $z^8$ . $z^8 = \left[ \sqrt{2} \left( \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) \right) \right]^8$ $z^8 = (\sqrt{2})^8 \left( \cos(8 \times -\frac{\pi}{4}) + i\sin(8 \times -\frac{\pi}{4}) \right)$ [M1 for applying theorem] $(\sqrt{2})^8 = (2^{1/2})^8 = 2^4 = 16$ . $8 \times -\frac{\pi}{4} = -2\pi$ . So, $z^8 = 16 (\cos(-2\pi) + i\sin(-2\pi))$ . Step 3: Convert back to Cartesian form. We evaluate the trigonometric functions: $\cos(-2\pi) = 1$ $\sin(-2\pi) = 0$ Therefore, $z^8 = 16(1 + i \cdot 0) = 16$ . [A1] The final answer is 16.

✓ FULLY WORKED

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEAComplex numbers

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 12

What is the imaginary unit, $i$ ?

TAP OR SPACE TO FLIP

ANSWER

$i$ is defined as the square root of -1, so $i^2 = -1$ . It allows us to find solutions to equations like $x^2 + 4 = 0$ .

TAP OR SPACE TO FLIP BACK

1 / 12 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
The modulus is always non-negative, as it represents a distance.
✓
The argument must be calculated carefully. Always sketch the point on an Argand diagram to ensure you find the angle in the correct quadrant.
✓
The principal argument is the unique value of $\theta$ such that $-\pi < \theta \le \pi$ . This is the standard range required in Cambridge exams unless specified otherwise.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Practice Complex Number Problems

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

Frequently asked

✓

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Practice Complex Number Problems on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS3.8 Differential equations NEXT →4.1 Forces and equilibrium

Complex numbers

BY THE END, YOU CAN…

1Understand and use the Cartesian form (x + iy) and polar form (r(cos θ + i sin θ) = re^(iθ)) of a complex number.
2Represent complex numbers on an Argand diagram and calculate their modulus and argument.
3Perform operations (multiplication, division, powers) on complex numbers in different forms.
4Apply De Moivre's theorem to find powers of complex numbers and solve related problems.

In simple terms

A friendly intro before the formal notes — no formulas yet.

Numbers in 2D

THINK OF IT LIKE…

1
A complex number z = x + iy is plotted as the point (x, y) in the Argand diagram, with a real (x) and imaginary (y) axis.
2
The modulus |z| is the distance from the origin, calculated as √(x² + y²). The argument, arg z, is the angle from the positive real axis.
3
The polar form uses the modulus (r) and argument (θ) to write z = r(cos θ + i sin θ), which simplifies to the exponential form z = re^(iθ).
4
De Moivre’s theorem provides a powerful shortcut for powers: (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ. This means to raise a complex number to a power, you raise the modulus to that power and multiply the argument.

Explore the concept

Use the live diagram and synced steps — play it or tap a step card to walk through.

Key formulas

Tap any symbol to reveal exactly what it means and its units.

z = x + iy

Tap a symbol — great for exam definitions

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

Tap a symbol — great for exam definitions

$[r(\cos\theta + i\sin\theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta))$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

Cartesian Form and the Argand Diagram

Modulus and Argument

For $z = x + iy$ : Modulus: $|z| = r = \sqrt{x^2 + y^2}$ Argument: $\tan(\theta) = \frac{y}{x}$

The modulus is always non-negative, as it represents a distance.
The argument must be calculated carefully. Always sketch the point on an Argand diagram to ensure you find the angle in the correct quadrant.
The principal argument is the unique value of $\theta$ such that $-\pi < \theta \le \pi$ . This is the standard range required in Cambridge exams unless specified otherwise.

Polar and Exponential Forms

Polar Form: $z = r(\cos\theta + i\sin\theta)$ Exponential Form: $z = re^{i\theta}$

Rules for Operations: If $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$ : $z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}$ $\frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)}$

EXAM TIP

De Moivre's Theorem

De Moivre's Theorem: $[r(\cos\theta + i\sin\theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta))$

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

Let the complex number $u = -\sqrt{3} + i$ . (i) Find the modulus and argument of $u$ . (ii) Express $u$ in exponential form.

1
Given $u = -\sqrt{3} + i$ . This is in the form $x+iy$ with $x = -\sqrt{3}$ and $y = 1$ . (i) Modulus and Argument First, calculate the modulus: $|u| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2$ . [M1, A1] Next, find the argument $\theta$ . Let's first find the related acute angle $\alpha$ : $\alpha = \tan^{-1} \left| \frac{y}{x} \right| = \tan^{-1} \left| \frac{1}{-\sqrt{3}} \right| = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6}$ . [M1] Now, sketch an Argand diagram. The point $(-\sqrt{3}, 1)$ is in the second quadrant. For the second quadrant, the principal argument is $\theta = \pi - \alpha$ . $\arg(u) = \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ . [A1] (ii) Exponential Form The exponential form is $z = re^{i\theta}$ . We have $r=2$ and $\theta = \frac{5\pi}{6}$ . So, $u = 2e^{i(5\pi/6)}$ . [B1 ft]

✓ FULLY WORKED

EXAMPLE 2

Use De Moivre's theorem to find $(1 - i)^8$ .

1
To use De Moivre's theorem, we must first convert $z = 1 - i$ into polar or exponential form. Step 1: Find modulus and argument of $z=1-i$ . $x=1, y=-1$ . Modulus: $|z| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$ . [M1] Argument: The point $(1, -1)$ is in the fourth quadrant. $\theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4}$ . [M1] So, $z = \sqrt{2} \left( \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) \right)$ . Step 2: Apply De Moivre's Theorem. We want to find $z^8$ . $z^8 = \left[ \sqrt{2} \left( \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) \right) \right]^8$ $z^8 = (\sqrt{2})^8 \left( \cos(8 \times -\frac{\pi}{4}) + i\sin(8 \times -\frac{\pi}{4}) \right)$ [M1 for applying theorem] $(\sqrt{2})^8 = (2^{1/2})^8 = 2^4 = 16$ . $8 \times -\frac{\pi}{4} = -2\pi$ . So, $z^8 = 16 (\cos(-2\pi) + i\sin(-2\pi))$ . Step 3: Convert back to Cartesian form. We evaluate the trigonometric functions: $\cos(-2\pi) = 1$ $\sin(-2\pi) = 0$ Therefore, $z^8 = 16(1 + i \cdot 0) = 16$ . [A1] The final answer is 16.

✓ FULLY WORKED

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEAComplex numbers

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Try to recall each definition before you reveal it.

Quick check

Answer in your head first — then tap to check. No pressure.

Revision flashcards

Flip the card. Test yourself before the exam.

QUESTION · 1 / 12

What is the imaginary unit, $i$ ?

TAP OR SPACE TO FLIP

ANSWER

$i$ is defined as the square root of -1, so $i^2 = -1$ . It allows us to find solutions to equations like $x^2 + 4 = 0$ .

TAP OR SPACE TO FLIP BACK

1 / 12 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

✓
The modulus is always non-negative, as it represents a distance.
✓
The argument must be calculated carefully. Always sketch the point on an Argand diagram to ensure you find the angle in the correct quadrant.
✓
The principal argument is the unique value of $\theta$ such that $-\pi < \theta \le \pi$ . This is the standard range required in Cambridge exams unless specified otherwise.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Real past paper[0]

Practice Complex Number Problems

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

Frequently asked

✓

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do Practice Complex Number Problems on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS3.8 Differential equations NEXT →4.1 Forces and equilibrium

Revise Complex numbers for Cambridge 9709 (3.9)

In simple terms

Explore the concept

Key formulas

Full topic notes

Cartesian Form and the Argand Diagram

Modulus and Argument

Polar and Exponential Forms

De Moivre's Theorem

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Practice Complex Number Problems

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

In simple terms

Explore the concept

Key formulas

Full topic notes

Cartesian Form and the Argand Diagram

Modulus and Argument

Polar and Exponential Forms

De Moivre's Theorem

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Practice Complex Number Problems

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Complex numbers