Further Complex Numbers

Q: What principal argument convention does Edexcel use?

Edexcel strictly uses the convention for all complex number questions. Always adjust arguments to this range before submitting final answers.

Q: Do I need to prove De Moivre's theorem in the exam?

Yes, you may be asked to prove De Moivre's theorem for all integer using induction. The formula booklet only states the result, not the proof, so you must recall the full inductive steps for positive, zero, and negative .

Q: Are CAS calculators allowed for FP2 papers?

No, CAS (Computer Algebra System) calculators are strictly forbidden for all Edexcel IAL Further Maths papers, including FP2. Standard scientific calculators are permitted.

Edexcel International A-Level Further Mathematics· FP2 3.1-3.4· 35 min read

1. Euler's Relation and Trigonometric Exponential Forms★★☆☆☆⏱ 8 min

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📘 Definition

Euler's Relation

$e^{i θ} = cos θ + i sin θ$ for all real $θ$ . This identity connects exponential functions to trigonometric functions for complex numbers.

Example:

For $θ = π$ , this gives Euler's identity: $e^{iπ} + 1 = 0$ .

From Euler's relation, you can derive two key identities for cosine and sine in terms of complex exponentials, which you must recall for the exam:

\cos\theta = \frac{1}{2}\left(e^{i\theta} + e^{-i\theta}\right)

\sin\theta = \frac{1}{2i}\left(e^{i\theta} - e^{-i\theta}\right)

📐 Worked Example

Express $cos^{3} θ$ in terms of multiple angles using the exponential form of cosine.

1
Start with the exponential identity for cosine, then expand the cube:
$\cos^3\theta = \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^3$
2
Expand the numerator using the binomial theorem:
$= \frac{e^{i3\theta} + 3e^{i\theta} + 3e^{-i\theta} + e^{-i3\theta}}{8}$
3
Group terms to match the exponential cosine identity structure:
$= \frac{1}{4}\left(\frac{e^{i3\theta} + e^{-i3\theta}}{2}\right) + \frac{3}{4}\left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)$
4
Substitute back to standard cosine form to get the final result:
$= \frac{1}{4}\cos3\theta + \frac{3}{4}\cos\theta$

Exam tip:

Exponential trig identities save significant time on multiple-angle identity questions compared to repeated double-angle rule application.

2. De Moivre's Theorem: Proof and Applications★★★☆☆⏱ 10 min

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📘 Definition

De Moivre's Theorem

For any integer $n$ and real $θ$ , $[r (cos θ + i sin θ)]^{n} = r^{n} (cos n θ + i sin n θ)$ . In exponential form, this simplifies to $(r e^{i θ})^{n} = r^{n} e^{in θ}$ .

🔬 Derivation

Goal:

Prove De Moivre's theorem for all integer values of $n$

Starting from:

Euler's relation and mathematical induction

1
Base case 1: $n = 1$ : $[r (cos θ + i sin θ)]^{1} = r (cos 1 θ + i sin 1 θ)$ , which is trivially true.
2
Inductive step: Assume the theorem holds for $n = k$ (positive integer): $[r (cos θ + i sin θ)]^{k} = r^{k} (cos k θ + i sin k θ)$ .
3
For $n = k + 1$ , multiply the $n = k$ result by $r (cos θ + i sin θ)$ and apply angle addition formulae:
$[r(\cos\theta + i\sin\theta)]^{k+1} = r^k(\cos k\theta + i\sin k\theta) \times r(\cos\theta + i\sin\theta) = r^{k+1}(\cos(k+1)\theta + i\sin(k+1)\theta)$
4
Base case 2: $n = 0$ : Left side = 1, right side = $r^{0} (cos 0 + i sin 0) = 1$ , so true for $n = 0$ .
5
Negative $n$ : Let $n = - m$ where $m$ is a positive integer. Use the fact that $\frac{1}{c o s m θ + i s i n m θ} = cos (- m θ) + i sin (- m θ)$ to show the theorem holds for all negative $n$ .

Result:

De Moivre's theorem is proven for all integer values of $n$ .

📐 Worked Example

Find all 4th roots of $z = 16 (cos \frac{2 π}{3} + i sin \frac{2 π}{3})$ .

1
Use De Moivre's theorem for roots: nth roots of $r (cos θ + i sin θ)$ have modulus $r^{1/ n}$ and arguments $\frac{θ + 2 π k}{n}$ for $k = 0, 1, ..., n - 1$ .
2
Calculate modulus of the roots: $1 6^{1/4} = 2$ .
3
Calculate the set of valid arguments for $k = 0, 1, 2, 3$ :
$\frac{2\pi/3 + 2\pi k}{4} = \frac{\pi}{6} + \frac{\pi k}{2}$
4
List the 4 distinct roots:
$2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right), 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right), 2\left(\cos\left(-\frac{5\pi}{6}\right) + i\sin\left(-\frac{5\pi}{6}\right)\right), 2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)$

Exam tip:

Always present nth roots with principal arguments (in the range $- π < θ \leq π$ ) unless explicitly instructed otherwise.

3. Loci and Regions in the Argand Diagram★★★☆☆⏱ 9 min

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$∣ z - a ∣ = b$ : Circle with centre at complex number $a$ , radius $b$
$∣ z - a ∣ = k ∣ z - b ∣$ ( $k \neq = 1$ ): Apollonius circle, all points $z$ with distance ratio $k$ from $a$ and $b$
$ar g (z - a) = β$ : Half-line starting at $a$ , making angle $β$ with the positive real axis, excluding the point $a$ itself
$ar g (\frac{z - a}{z - b}) = β$ : Arc of a circle passing through $a$ and $b$ , where the angle between lines from $z$ to $a$ and $z$ to $b$ is $β$
Regions are inequalities of these forms, e.g. $∣ z - a ∣ \leq b$ is the interior and boundary of the circle, $∣ z - a ∣ \leq ∣ z - b ∣$ is the half-plane closer to $a$ than $b$

📐 Worked Example

Sketch the locus defined by $ar g (\frac{z - 2 i}{z + 3}) = \frac{π}{4}$ , and state the coordinates of the centre of the corresponding circle.

1
Identify the two fixed points: $a = 2 i$ (coordinates $(0, 2)$ ) and $b = - 3$ (coordinates $(- 3, 0)$ ).
2
The locus is the arc of the circle passing through $(0, 2)$ and $(- 3, 0)$ where the angle subtended by the chord joining these two points is $\frac{π}{4}$ , above the chord.
3
The centre lies on the perpendicular bisector of the chord between $(- 3, 0)$ and $(0, 2)$ . The midpoint is $(- 1.5, 1)$ , slope of the chord is $\frac{2}{3}$ , so the slope of the perpendicular bisector is $- \frac{3}{2}$ .
4
Use the circle theorem that the central angle is twice the inscribed angle, so the central angle is $\frac{π}{2}$ . The centre lies on the perpendicular bisector at distance $\frac{13}{2}$ from the midpoint $(- 1.5, 1)$ , giving the centre $(- \frac{1}{2}, - \frac{1}{2})$ with radius $\frac{13}{2}$ .

Exam tip:

Always label fixed points and key angles on locus sketches to get full method marks, even if your sketch is not perfectly to scale.

4. z-plane to w-plane Transformations★★★★☆⏱ 8 min

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📘 Definition

Möbius Transformation

A transformation of the form $w = \frac{a z + b}{cz + d}$ where $a, b, c, d$ are complex constants and $a d - b c \neq = 0$ . These map lines and circles in the z-plane to lines or circles in the w-plane.

📐 Worked Example

The transformation $w = \frac{z + 1}{z - i}$ maps the locus $∣ z ∣ = 1$ in the z-plane to a line in the w-plane. Find the equation of this line.

1
Rearrange the transformation to make $z$ the subject:
$w(z - i) = z + 1 \implies wz - wi = z + 1 \implies z(w - 1) = 1 + wi \implies z = \frac{1 + wi}{w - 1}$
2
Substitute into the original locus equation $∣ z ∣ = 1$ :
$\left|\frac{1 + wi}{w - 1}\right| = 1$
3
Use the modulus property $\frac{A}{B} = \frac{∣ A ∣}{∣ B ∣}$ to simplify:
$|1 + wi| = |w - 1|$
4
Let $w = u + i v$ (real $u, v$ ) and substitute into the equation:
$|1 + i(u + iv)| = |u + iv - 1| \implies |1 - v + iu| = |(u - 1) + iv|$
5
Equate squared moduli and simplify:
$(1 - v)^2 + u^2 = (u - 1)^2 + v^2 \implies 1 - 2v + v^2 + u^2 = u^2 - 2u + 1 + v^2 \implies u = v$

Exam tip:

When asked for the equation of a transformed locus, always express your final answer in terms of real coordinates $u$ and $v$ (where $w = u + i v$ ) unless asked for complex form.

5. Common Pitfalls

Wrong move:

Using the principal argument range $0 \leq ar g (z) < 2 π$ instead of Edexcel's required $- π < ar g (z) \leq π$

Why:

Edexcel explicitly specifies the negative to positive pi convention, so answers outside this range will lose accuracy marks.

Correct move:

Always adjust arguments by adding or subtracting $2 π$ to bring them into the $(- π, π]$ range before submitting final answers.

Wrong move:

Drawing $ar g (z - a) = β$ as a full infinite line instead of a half-line

Why:

The argument is only defined for points in the direction of $β$ from $a$ , and the point $a$ itself is excluded (argument is undefined there).

Correct move:

Sketch only the part of the line at angle $β$ from $a$ in the correct direction, and mark $a$ with an open circle to show it is excluded.

Wrong move:

Forgetting to divide $2 π k$ by $n$ when calculating nth roots of a complex number

Why:

The arguments of nth roots are spaced by $2 π / n$ , so adding full $2 π$ increments will repeat the same root instead of giving distinct roots.

Correct move:

For $n$ distinct roots, use arguments $\frac{θ + 2 π k}{n}$ for $k = 0, 1, ..., n - 1$ .

Wrong move:

Assuming Möbius transformations always map circles to circles

Why:

If the original circle in the z-plane passes through the pole of the transformation (the point $z = - d / c$ where the denominator is zero), it maps to a line in the w-plane, not a circle.

Correct move:

Check if the pole lies on the original locus first to determine if the transformed locus is a line or circle.

Wrong move:

Only proving De Moivre's theorem for positive integer $n$ and neglecting $n = 0$ and negative $n$

Why:

Exam questions often ask for proof for all integers $n$ , so missing the zero and negative cases will lose approximately half the available marks.

Correct move:

Always include base cases for $n = 1$ and $n = 0$ , then extend to negative $n$ by taking reciprocals of positive $n$ results.

6. Quick Reference Cheatsheet

Concept	Key Formula/Rule	Exam Reminder
Euler's Relation	$e^{i θ} = cos θ + i sin θ$	Derive $cos θ = \frac{e ^{i θ} + e ^{- i θ}}{2}$ and $sin θ = \frac{e ^{i θ} - e ^{- i θ}}{2 i}$ if needed
De Moivre's Theorem	$[r (cos θ + i sin θ)]^{n} = r^{n} (cos n θ + i sin n θ)$	Prove via induction for all integers n; roots spaced by $2 π / n$
$∣ z - a ∣ = b$	Circle, centre $a$ , radius $b$	Region $∣ z - a ∣ \leq b$ includes boundary
$∣ z - a ∣ = k ∣ z - b ∣$ ( $k \neq = 1$ )	Apollonius circle	If $k = 1$ , it is the perpendicular bisector of $a$ and $b$
$ar g (z - a) = β$	Half-line from $a$ , angle $β$ to real axis	Exclude point $a$ (open circle on sketch)
Möbius Transformation	$w = \frac{a z + b}{cz + d}$	Rearrange to make $z$ subject, substitute original locus equation

7. Frequently Asked

What principal argument convention does Edexcel use?

Edexcel strictly uses the convention $- π < ar g (z) \leq π$ for all complex number questions. Always adjust arguments to this range before submitting final answers.

Do I need to prove De Moivre's theorem in the exam?

Yes, you may be asked to prove De Moivre's theorem for all integer $n$ using induction. The formula booklet only states the result, not the proof, so you must recall the full inductive steps for positive, zero, and negative $n$ .

Are CAS calculators allowed for FP2 papers?

No, CAS (Computer Algebra System) calculators are strictly forbidden for all Edexcel IAL Further Maths papers, including FP2. Standard scientific calculators are permitted.

Going deeper

What's Next

Now that you have mastered Further Complex Numbers for Edexcel IAL FP2, you are ready to move on to the next core FP2 topics, starting with second-order differential equations, which often use complex number methods for solving higher-order linear equations. You should also practice full past paper questions on this topic to familiarize yourself with the mix of proofs, calculations, and sketching questions that appear annually. Ensure you can quickly recall the standard loci and transformation steps to save time in your 1h30 exam.