Complex numbers

Q: What is Euler's formula and the resulting exponential form of a complex number?

Euler's formula states that \(e^{i\theta} = \cos\theta + i\sin\theta\), which gives the exponential form \(z = r e^{i\theta}\).

Q: How are multiplication and division performed on complex numbers in exponential form?

For multiplication, multiply moduli and add arguments: \(z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}\). For division, divide moduli and subtract arguments: \(\frac{z_1}{z_2} = \left(\frac{r_1}{r_2}\right) e^{i(\theta_1 - \theta_2)}\).

Q: What is De Moivre's Theorem?

For any real number \(n\), \([r(\cos\theta + i\sin\theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta))\), or simply \((\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)\) when \(r = 1\).

Q: How do you express \(\cos(n\theta)\) or \(\sin(n\theta)\) in powers of \(\cos\theta\) and \(\sin\theta\)?

Expand \((\cos\theta + i\sin\theta)^n\) using the Binomial Theorem and equate real parts to find \(\cos(n\theta)\) or imaginary parts to find \(\sin(n\theta)\).

Q: What are the key identities used to express powers \(\cos^n\theta\) and \(\sin^n\theta\) in terms of multiple angles?

\(z^n + \frac{1}{z^n} = 2\cos(n\theta)\) and \(z^n - \frac{1}{z^n} = 2i\sin(n\theta)\), where \(z = \cos\theta + i\sin\theta\).

Q: What is the general method for finding the \(n\)-th roots of a complex number \(w\)?

Write \(w = R e^{i(\phi + 2k\pi)}\), then apply \(z = R^{\frac{1}{n}} e^{i\left(\frac{\phi + 2k\pi}{n}\right)}\) and substitute \(n\) consecutive integers for \(k\) (e.g., \(k = 0, 1, \dots, n-1\)).

Q: What are the \(n\)-th roots of unity and their key sum property?

They are the solutions to \(z^n = 1\), given by \(1, \omega, \omega^2, \dots, \omega^{n-1}\) where \(\omega = e^{i\frac{2\pi}{n}}\), and their sum is always zero: \(1 + \omega + \omega^2 + \dots + \omega^{n-1} = 0\).

Q: How is the \(C + iS\) method used to sum trigonometric series?

Form \(C + iS\) to create a geometric series in powers of \(e^{i\theta}\), find its sum using the geometric series formula, and equate real and imaginary parts to obtain \(C\) and \(S\).

Q: What geometric locus is represented by \(|z - z_1| = r\)?

A circle with centre at \(z_1\) and radius \(r\).

Q: What geometric locus is represented by \(|z - z_1| = |z - z_2|\)?

The perpendicular bisector of the line segment joining the points \(z_1\) and \(z_2\).

Further Mathematics 2330 · CCEA A Level · 10 min read · Updated: 31 Aug 2026

Welcome to A2 Complex Numbers!

Welcome to one of the most elegant and powerful topics in Further Mathematics! In AS Further Maths, you were introduced to the imaginary unit \(i = \sqrt{-1}\), Cartesian form \(z = x + iy\), and the modulus-argument form \(z = r(\cos\theta + i\sin\theta)\). Now, in A2 Unit 1, we unlock the true superpowers of complex numbers.

Complex numbers are not just abstract mathematical puzzles; they are vital in the real world for electrical engineering (modelling alternating current circuits), quantum mechanics, aerodynamics, and computer graphics.

Don't worry if this chapter seems intimidating at first! We will break down every concept into step-by-step chunks, share memory tricks, and highlight common pitfalls so you can master this topic with confidence.

1. Quick Refresh & Euler's Formula

Before diving into new techniques, let's briefly review the different ways we can write a complex number \(z\):

1. Cartesian Form: \(z = x + iy\), where \(x = \text{Re}(z)\) and \(y = \text{Im}(z)\).
2. Modulus-Argument (Polar) Form: \(z = r(\cos\theta + i\sin\theta)\), where \(r = |z| = \sqrt{x^2 + y^2}\) is the distance from the origin, and \(\theta = \arg(z)\) is the angle measured anticlockwise from the positive real axis, with principal argument \(-\pi < \theta \le \pi\).

Euler's Formula and Exponential Form

Leonhard Euler discovered an incredible connection between trigonometry and complex exponentials:

\(e^{i\theta} = \cos\theta + i\sin\theta\)

This allows us to write any complex number in exponential form:

\(z = r e^{i\theta}\)

Did you know? If you substitute \(\theta = \pi\), you get \(e^{i\pi} = \cos\pi + i\sin\pi = -1 + 0i\), which leads to Euler's Identity: \(e^{i\pi} + 1 = 0\). This famous equation links the five fundamental mathematical constants: \(0, 1, e, i,\) and \(\pi\)!

Multiplying and Dividing in Exponential Form

Working with exponentials makes multiplication and division remarkably straightforward because standard index laws apply:

Multiplication: \(z_1 z_2 = (r_1 e^{i\theta_1})(r_2 e^{i\theta_2}) = r_1 r_2 e^{i(\theta_1 + \theta_2)}\)
Rule: Multiply the moduli, add the arguments!

Division: \(\frac{z_1}{z_2} = \frac{r_1 e^{i\theta_1}}{r_2 e^{i\theta_2}} = \left(\frac{r_1}{r_2}\right) e^{i(\theta_1 - \theta_2)}\)
Rule: Divide the moduli, subtract the arguments!

Key Takeaway: Exponential form \(z = r e^{i\theta}\) is completely equivalent to polar form \(z = r(\cos\theta + i\sin\theta)\), but makes multiplication and division much faster.

2. De Moivre's Theorem

De Moivre's Theorem is the central pillar of A2 Complex Numbers. It states that for any real number \(n\):

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

When \(r = 1\), this simplifies to:

\((\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)\)

Proof by Mathematical Induction (for positive integers \(n\))

Exam questions often ask for this proof, so let's walk through it step-by-step:

Base Step: For \(n = 1\), \((\cos\theta + i\sin\theta)^1 = \cos(1\theta) + i\sin(1\theta)\). The statement holds for \(n = 1\).

Inductive Step: Assume the result is true for \(n = k\), so \((\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta)\).
Now consider \(n = k + 1\):
\((\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)^k (\cos\theta + i\sin\theta)\)
\(= (\cos(k\theta) + i\sin(k\theta))(\cos\theta + i\sin\theta)\)
\(= (\cos(k\theta)\cos\theta - \sin(k\theta)\sin\theta) + i(\sin(k\theta)\cos\theta + \cos(k\theta)\sin\theta)\)
Using compound angle formulae:
\(= \cos(k\theta + \theta) + i\sin(k\theta + \theta)\)
\(= \cos((k+1)\theta) + i\sin((k+1)\theta)\)

Conclusion: Since the statement is true for \(n = 1\), and if true for \(n = k\) it is also true for \(n = k + 1\), by mathematical induction it is true for all integers \(n \ge 1\).

Key Takeaway: Raising a complex number to the power \(n\) means raising its modulus to the power \(n\) and multiplying its argument by \(n\).

3. Trigonometric Applications of De Moivre's Theorem

De Moivre's theorem gives us a bridge between trigonometry and algebra. There are two standard exam question types you must master.

Type 1: Expressing \(\cos(n\theta)\) and \(\sin(n\theta)\) in Powers of \(\cos\theta\) and \(\sin\theta\)

Method:
1. Write down \((\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)\).
2. Expand the left-hand side using the Binomial Theorem.
3. Equate real parts to find \(\cos(n\theta)\), or equate imaginary parts to find \(\sin(n\theta)\).

Example: Express \(\cos(3\theta)\) in terms of \(\cos\theta\).
\(\cos(3\theta) + i\sin(3\theta) = (\cos\theta + i\sin\theta)^3\)
\(= \cos^3\theta + 3\cos^2\theta(i\sin\theta) + 3\cos\theta(i\sin\theta)^2 + (i\sin\theta)^3\)
\(= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta\)
Equating real parts:
\(\cos(3\theta) = \cos^3\theta - 3\cos\theta\sin^2\theta\)
Replace \(\sin^2\theta\) with \(1 - \cos^2\theta\):
\(\cos(3\theta) = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) = 4\cos^3\theta - 3\cos\theta\).

Type 2: Expressing \(\cos^n\theta\) or \(\sin^n\theta\) as Multiple Angles

Let \(z = \cos\theta + i\sin\theta\). Then \(\frac{1}{z} = z^{-1} = \cos(-\theta) + i\sin(-\theta) = \cos\theta - i\sin\theta\).

This gives two essential relationships to memorise:

Cosine relationship: \(z + \frac{1}{z} = 2\cos\theta\) and \(z^n + \frac{1}{z^n} = 2\cos(n\theta)\)
Sine relationship: \(z - \frac{1}{z} = 2i\sin\theta\) and \(z^n - \frac{1}{z^n} = 2i\sin(n\theta)\)

Memory Aid: Plus gives Cosine (both letters curved/friendly), Minus gives Sine with an \(i\) (don't forget the \(i\) on sine!).

Example: Express \(\cos^4\theta\) in terms of multiple angles.
Start with \((2\cos\theta)^4 = \left(z + \frac{1}{z}\right)^4\):
\(16\cos^4\theta = z^4 + 4z^3\left(\frac{1}{z}\right) + 6z^2\left(\frac{1}{z^2}\right) + 4z\left(\frac{1}{z^3}\right) + \frac{1}{z^4}\)
Group terms with matching powers:
\(16\cos^4\theta = \left(z^4 + \frac{1}{z^4}\right) + 4\left(z^2 + \frac{1}{z^2}\right) + 6\)
Substitute \(z^n + \frac{1}{z^n} = 2\cos(n\theta)\):
\(16\cos^4\theta = 2\cos(4\theta) + 4(2\cos(2\theta)) + 6\)
\(16\cos^4\theta = 2\cos(4\theta) + 8\cos(2\theta) + 6\)
Divide by 16:
\(\cos^4\theta = \frac{1}{8}\cos(4\theta) + \frac{1}{2}\cos(2\theta) + \frac{3}{8}\)

Why do this? This makes integrating powers of trigonometric functions like \(\int \cos^4\theta \, d\theta\) remarkably simple!

Key Takeaway: Use the binomial expansion of \((\cos\theta + i\sin\theta)^n\) for \(\cos(n\theta)\), and use expansions of \(\left(z \pm \frac{1}{z}\right)^n\) for powers like \(\cos^n\theta\) or \(\sin^n\theta\).

4. Finding the \(n\)-th Roots of a Complex Number

Solving equations of the form \(z^n = w\) (where \(w\) is a given complex number) is a core skill.

Step-by-Step Method

Step 1: Write the complex number \(w\) in modulus-argument form with a general argument:
\(w = R(\cos(\phi + 2k\pi) + i\sin(\phi + 2k\pi))\) or \(w = R e^{i(\phi + 2k\pi)}\), where \(k \in \mathbb{Z}\).

Step 2: Apply De Moivre's Theorem to find \(z = w^{\frac{1}{n}}\):
\(z = R^{\frac{1}{n}} \left( \cos\left(\frac{\phi + 2k\pi}{n}\right) + i\sin\left(\frac{\phi + 2k\pi}{n}\right) \right)\)

Step 3: Substitute \(n\) consecutive integer values for \(k\) (usually \(k = 0, 1, 2, \dots, n-1\) or values chosen to keep arguments in the range \(-\pi < \theta \le \pi\)).

Geometric Interpretation on the Argand Diagram

The \(n\) roots of any complex number have remarkable geometric symmetry:
- All roots lie on a circle centered at the origin with radius \(r = R^{\frac{1}{n}}\).
- The roots are equally spaced around the circle, separated by an angle of \(\frac{2\pi}{n}\) radians (\(\frac{360^\circ}{n}\)).
- Joining the points forms a regular \(n\)-sided polygon (e.g., 3 roots form an equilateral triangle, 4 roots form a square, 5 roots form a regular pentagon).

Roots of Unity

The solutions to \(z^n = 1\) are called the \(n\)-th roots of unity.
Since \(1 = 1 e^{i(0 + 2k\pi)}\), the roots are given by:
\(z = e^{i\frac{2k\pi}{n}} = \cos\left(\frac{2k\pi}{n}\right) + i\sin\left(\frac{2k\pi}{n}\right)\) for \(k = 0, 1, \dots, n-1\).

If we let \(\omega = e^{i\frac{2\pi}{n}}\), the roots are \(1, \omega, \omega^2, \dots, \omega^{n-1}\).
Crucial Property: The sum of all \(n\)-th roots of unity is always zero:
\(1 + \omega + \omega^2 + \dots + \omega^{n-1} = 0\)

Key Takeaway: An equation \(z^n = w\) always has exactly \(n\) distinct roots. On an Argand diagram, they form the vertices of a regular polygon.

5. Summation of Trigonometric Series (\(C + iS\) Method)

We can sum series involving sines and cosines using geometric series of complex numbers.

The Method

Suppose you need to find the sums:
\(C = 1 + \cos\theta + \cos(2\theta) + \dots + \cos(n\theta)\)
\(S = \sin\theta + \sin(2\theta) + \dots + \sin(n\theta)\)

Step 1: Form the combined series \(C + iS\):
\(C + iS = 1 + (\cos\theta + i\sin\theta) + (\cos(2\theta) + i\sin(2\theta)) + \dots + (\cos(n\theta) + i\sin(n\theta))\)
\(= 1 + e^{i\theta} + e^{i2\theta} + \dots + e^{in\theta}\)

Step 2: Notice that this is a geometric progression with first term \(a = 1\) and common ratio \(z = e^{i\theta}\). Use the formula for the sum of a geometric series:
\(S_N = \frac{a(1 - z^N)}{1 - z}\)

Step 3: Simplify the resulting complex expression by multiplying the numerator and denominator by the complex conjugate or by factoring out half-angles (e.g., \(1 - e^{i\theta} = e^{i\theta/2}(e^{-i\theta/2} - e^{i\theta/2}) = -2i e^{i\theta/2}\sin\left(\frac{\theta}{2}\right)\)).

Step 4: Equate the real part to find \(C\) and the imaginary part to find \(S\).

Key Takeaway: Combine trigonometric series into \(C + iS\), apply standard geometric series formulae, and equate real and imaginary parts.

6. Loci and Regions in the Complex Plane

A locus (plural: loci) is a set of points \(z = x + iy\) satisfying a given geometric condition.

1. Circle: \(|z - z_1| = r\)

- Represents the set of points whose distance from a fixed point \(z_1\) is equal to \(r\).
- Geometry: A circle with centre \(z_1\) and radius \(r\).
- Example: \(|z - (3 + 2i)| = 5\) is a circle centred at \((3, 2)\) with radius \(5\).
- Region: \(|z - z_1| \le r\) represents the circle plus all points inside it.

2. Perpendicular Bisector: \(|z - z_1| = |z - z_2|\)

- Represents the set of points that are equidistant from two fixed points \(z_1\) and \(z_2\).
- Geometry: The perpendicular bisector of the line segment joining \(z_1\) and \(z_2\).
- Region: \(|z - z_1| < |z - z_2|\) is the region (half-plane) strictly closer to \(z_1\) than to \(z_2\).

3. Half-Line (Ray): \(\arg(z - z_1) = \alpha\)

- Represents the set of points \(z\) such that the vector from \(z_1\) to \(z\) makes an angle of \(\alpha\) with the horizontal positive real direction.
- Geometry: A half-line (or ray) starting at \(z_1\) (not including \(z_1\) itself, so draw an open circle at \(z_1\)) making an angle \(\alpha\) with the positive horizontal line.
- Region: \(\alpha \le \arg(z - z_1) \le \beta\) represents the sector between two rays emanating from \(z_1\).

Quick Loci Reference Table

Equation: \(|z - z_1| = r\) → Locus: Circle, centre \(z_1\), radius \(r\)
Equation: \(|z - z_1| = |z - z_2|\) → Locus: Perpendicular bisector of segment \(z_1 z_2\)
Equation: \(\arg(z - z_1) = \alpha\) → Locus: Half-line starting at \(z_1\) at angle \(\alpha\)

Key Takeaway: Always rewrite expressions as \(|z - z_1|\) with a minus sign inside to correctly identify the centre or starting point \(z_1\)!

7. Common Mistakes to Avoid

1. Sign errors when finding centres: For \(|z + 2 - 3i| = 4\), rewrite it as \(|z - (-2 + 3i)| = 4\). The centre is \(-2 + 3i\), not \(2 - 3i\).
2. Forgetting the \(i\) in Sine expansions: Remember that \(z - \frac{1}{z} = 2i\sin\theta\). When raising this to an even power, \((2i)^4 = 16i^4 = 16\), but for an odd power, \((2i)^3 = 8i^3 = -8i\). Watch those powers of \(i\)!
3. Forgetting open circles on rays: \(\arg(z - z_1) = \alpha\) does not include the endpoint \(z_1\) because \(\arg(0)\) is undefined. Always draw an open hollow circle at \(z_1\).
4. Missing roots: When solving \(z^n = w\), ensure you provide all \(n\) distinct solutions. Check that your arguments lie in the requested range (usually \(-\pi < \theta \le \pi\)).

8. Chapter Summary Checklist

Before sitting your exam, make sure you can:
• Convert seamlessly between Cartesian, Mod-Arg, and Exponential forms.
• State and prove De Moivre's Theorem for positive integers by induction.
• Use De Moivre's Theorem to express \(\cos(n\theta)\) and \(\sin(n\theta)\) as polynomials in \(\cos\theta\) and \(\sin\theta\).
• Use \(z + \frac{1}{z} = 2\cos\theta\) and \(z - \frac{1}{z} = 2i\sin\theta\) to express \(\cos^n\theta\) and \(\sin^n\theta\) in terms of multiple angles.
• Solve \(z^n = w\) and sketch the roots as a regular polygon on an Argand diagram.
• Sum trigonometric series using the \(C + iS\) method.
• Sketch and interpret loci and regions involving circles, perpendicular bisectors, and rays.

Quick check

Can you answer these now?

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

What is Euler's formula and the resulting exponential form of a complex number?

Euler's formula states that \(e^{i\theta} = \cos\theta + i\sin\theta\), which gives the exponential form \(z = r e^{i\theta}\).

How are multiplication and division performed on complex numbers in exponential form?

For multiplication, multiply moduli and add arguments: \(z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}\). For division, divide moduli and subtract arguments: \(\frac{z_1}{z_2} = \left(\frac{r_1}{r_2}\right) e^{i(\theta_1 - \theta_2)}\).

What is De Moivre's Theorem?

For any real number \(n\), \([r(\cos\theta + i\sin\theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta))\), or simply \((\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)\) when \(r = 1\).

How do you express \(\cos(n\theta)\) or \(\sin(n\theta)\) in powers of \(\cos\theta\) and \(\sin\theta\)?

Expand \((\cos\theta + i\sin\theta)^n\) using the Binomial Theorem and equate real parts to find \(\cos(n\theta)\) or imaginary parts to find \(\sin(n\theta)\).

What are the key identities used to express powers \(\cos^n\theta\) and \(\sin^n\theta\) in terms of multiple angles?

\(z^n + \frac{1}{z^n} = 2\cos(n\theta)\) and \(z^n - \frac{1}{z^n} = 2i\sin(n\theta)\), where \(z = \cos\theta + i\sin\theta\).

What is the general method for finding the \(n\)-th roots of a complex number \(w\)?

Write \(w = R e^{i(\phi + 2k\pi)}\), then apply \(z = R^{\frac{1}{n}} e^{i\left(\frac{\phi + 2k\pi}{n}\right)}\) and substitute \(n\) consecutive integers for \(k\) (e.g., \(k = 0, 1, \dots, n-1\)).

What are the \(n\)-th roots of unity and their key sum property?

They are the solutions to \(z^n = 1\), given by \(1, \omega, \omega^2, \dots, \omega^{n-1}\) where \(\omega = e^{i\frac{2\pi}{n}}\), and their sum is always zero: \(1 + \omega + \omega^2 + \dots + \omega^{n-1} = 0\).

How is the \(C + iS\) method used to sum trigonometric series?

Form \(C + iS\) to create a geometric series in powers of \(e^{i\theta}\), find its sum using the geometric series formula, and equate real and imaginary parts to obtain \(C\) and \(S\).

What geometric locus is represented by \(|z - z_1| = r\)?

A circle with centre at \(z_1\) and radius \(r\).

What geometric locus is represented by \(|z - z_1| = |z - z_2|\)?

The perpendicular bisector of the line segment joining the points \(z_1\) and \(z_2\).

What geometric locus is represented by \(\arg(z - z_1) = \alpha\)?

A half-line (ray) starting at \(z_1\) (with an open circle at \(z_1\)) making an angle \(\alpha\) with the positive real direction.

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