Notes · Further MathematicsUK · A-Levels

Complex numbers

Complex numbers extend the real numbers by adjoining a square root of $-1$ , and Further Mathematics develops their geometry on the Argand diagram in full. The central results are the modulus-argument and exponential forms, de Moivre's theorem (with its uses for trigonometric identities and roots), the roots of unity, and loci in the complex plane. Complex numbers tie together algebra, trigonometry and geometry, and reappear in the auxiliary equation of second-order differential equations.

5 sections~15 min reading time3 competenciesLevel Standard 2 · Advanced 3

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Exam profile

AO1 · Perform complex arithmetic and convert accurately between Cartesian, modulus-argument and exponential formsAO2 · Prove de Moivre's theorem by induction and derive trigonometric identities from itAO3 · Interpret, sketch and use loci and regions in the Argand diagram to solve geometric problems

Operators:expressfindsolveproveshow thatsketchhencededuce

basic level

AS Further Mathematics covers complex arithmetic, the Argand diagram, modulus-argument form and simple loci such as $|z-a|=r$ .

higher level

The full A-Level adds de Moivre's theorem, the $n$ th roots of a complex number and the roots of unity, exponential form, and the full range of loci including perpendicular bisectors and half-lines.

Depth

Text

Contents · 5 sections

Complex numbers

Complex arithmetic and the Argand diagram#

StandardAQA 7367 §3.2DfE Further Mathematics — Complex numbers

Addition of complex numbers as vectors

Fig. 1The sum

z+w

is the diagonal of the parallelogram built on the position vectors of

z

and

w

A complex number is written

z = a + b\mathrm{i}

, where

a = \operatorname{Re}(z)

is the real part,

b = \operatorname{Im}(z)

is the imaginary part, and

\mathrm{i}

satisfies

\mathrm{i}^2 = -1

. Addition and subtraction act componentwise, and multiplication is carried out by expanding and using

\mathrm{i}^2 = -1

(a+b\mathrm{i})(c+d\mathrm{i}) = (ac-bd) + (ad+bc)\mathrm{i}

. Division is made real in the denominator by multiplying top and bottom by the conjugate, exactly as one rationalises a surd.

The complex conjugate

\bar z = a - b\mathrm{i}

is fundamental. It satisfies

z\bar z = a^2 + b^2

, a non-negative real number, which is what makes division work:

\dfrac{1}{z} = \dfrac{\bar z}{z\bar z} = \dfrac{a - b\mathrm{i}}{a^2+b^2}

. The conjugate also respects the arithmetic operations —

\overline{z+w} = \bar z + \bar w

and

\overline{zw} = \bar z\,\bar w

— and a number is real precisely when

z = \bar z

. A key consequence, used when solving polynomials, is that the non-real roots of a polynomial with real coefficients occur in conjugate pairs.

The Argand diagram represents

z = a + b\mathrm{i}

as the point

(a, b)

, with the horizontal axis carrying the real part and the vertical axis the imaginary part. Crucially,

z

can equally be pictured as the position vector from the origin to that point, and this is why complex-number geometry is so powerful: addition of complex numbers is exactly vector addition, obeying the parallelogram law, and

z - w

is the vector that points from

w

z

Reading geometry off the diagram is a core skill. The distance of

z

from the origin is

\sqrt{a^2+b^2}

, and the distance between two complex numbers

z

and

w

|z - w|

. This single fact — that

|z-w|

is the distance between the points — underlies every locus later in the topic. Conjugation is reflection in the real axis, and multiplication by

-1

is a half-turn about the origin, so even elementary operations have clean geometric meanings.

z\bar z = (a+b\mathrm{i})(a-b\mathrm{i}) = a^2 + b^2 = |z|^2

Conjugate product

Multiplying by the conjugate gives a non-negative real number, the square of the modulus; this makes division possible.

\frac{z}{w} = \frac{z\,\bar w}{w\,\bar w} = \frac{z\,\bar w}{|w|^2}

Division by the conjugate

Multiply numerator and denominator by $\bar w$ to make the denominator real.

Worked example

Division and a conjugate root

(a) Express $\dfrac{5 + 5\mathrm{i}}{3 - 4\mathrm{i}}$ in the form $a + b\mathrm{i}$ . (b) Given that $2 - \mathrm{i}$ is a root of $z^2 + pz + q = 0$ with $p, q$ real, find $p$ and $q$ .

Multiply by the conjugate
$\dfrac{5+5\mathrm{i}}{3-4\mathrm{i}}\cdot\dfrac{3+4\mathrm{i}}{3+4\mathrm{i}} = \dfrac{(5+5\mathrm{i})(3+4\mathrm{i})}{9+16}$ .
Expand the numerator
$(5+5\mathrm{i})(3+4\mathrm{i}) = 15 + 20\mathrm{i} + 15\mathrm{i} + 20\mathrm{i}^2 = 15 + 35\mathrm{i} - 20 = -5 + 35\mathrm{i}$ .
$\frac{5+5\mathrm{i}}{3-4\mathrm{i}} = \frac{-5+35\mathrm{i}}{25} = -\tfrac{1}{5} + \tfrac{7}{5}\mathrm{i}$
Use the conjugate root
Since the coefficients are real, the other root is $\overline{2-\mathrm{i}} = 2+\mathrm{i}$ . Sum of roots $=(2-\mathrm{i})+(2+\mathrm{i}) = 4 = -p$ , so $p=-4$ .
Find q
Product of roots $=(2-\mathrm{i})(2+\mathrm{i}) = 4 - \mathrm{i}^2 = 5 = q$ .

Result: (a) $-\tfrac{1}{5} + \tfrac{7}{5}\mathrm{i}$ ; (b) $p = -4$ , $q = 5$ .

Typical mistakes

Sign slips with $\mathrm{i}^2 = -1$ , especially forgetting the sign change when squaring the imaginary part.
Assuming complex roots come in conjugate pairs even when the polynomial has non-real coefficients — the property requires real coefficients.

Active revision

Given $z = 3 + \mathrm{i}$ and $w = 1 + 2\mathrm{i}$ , find $z+w$ , $zw$ and $\dfrac{z}{w}$ in the form $a + b\mathrm{i}$ , and mark $z$ , $w$ and $z+w$ on an Argand diagram.

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Modulus-argument and exponential form#

StandardAQA 7367 §3.2DfE Further Mathematics — Complex numbers

Modulus and argument of z = 1 + √3 i

Fig. 2The modulus is the length of the position vector; the argument is the angle it makes with the positive real axis.

The modulus

|z| = r = \sqrt{a^2+b^2}

is the distance of

z

from the origin, and the argument

\arg z = \theta

is the angle the position vector makes with the positive real axis, measured anticlockwise and conventionally taken in the principal range

-\pi < \theta \leq \pi

. Then

a = r\cos\theta

and

b = r\sin\theta

, giving the modulus-argument form

z = r(\cos\theta + \mathrm{i}\sin\theta)

, often abbreviated

r\,\operatorname{cis}\theta

Finding the argument demands care over the quadrant. The equation

\tan\theta = b/a

has two solutions in a full turn, so you must use the signs of

a

and

b

to place

z

in the correct quadrant rather than blindly taking

\arctan(b/a)

. For example

z = -1 + \mathrm{i}

has

\tan\theta = -1

but lies in the second quadrant, so

\arg z = \tfrac{3\pi}{4}

, not

-\tfrac{\pi}{4}

. Sketching

z

before computing the argument prevents almost all such errors.

Modulus-argument form turns multiplication into a rotation-and-scaling. Multiplying two complex numbers multiplies their moduli and adds their arguments; dividing divides the moduli and subtracts the arguments. Thus

|zw| = |z||w|

and

\arg(zw) = \arg z + \arg w

. Geometrically, multiplying by

z = r\operatorname{cis}\theta

enlarges by factor

r

and rotates by angle

\theta

about the origin — a fact used constantly with de Moivre's theorem and with transformations.

Euler's relation

\mathrm{e}^{\mathrm{i}\theta} = \cos\theta + \mathrm{i}\sin\theta

packages this compactly:

z = r\mathrm{e}^{\mathrm{i}\theta}

is the exponential form. It makes the multiplication rule transparent, since

r_1\mathrm{e}^{\mathrm{i}\theta_1}\cdot r_2\mathrm{e}^{\mathrm{i}\theta_2} = r_1 r_2\,\mathrm{e}^{\mathrm{i}(\theta_1+\theta_2)}

by the ordinary law of indices. The special case

\mathrm{e}^{\mathrm{i}\pi} = -1

(Euler's identity) links the five most important constants of mathematics in a single equation.

z = r(\cos\theta + \mathrm{i}\sin\theta) = r\,\mathrm{e}^{\mathrm{i}\theta}, \qquad r = |z|,\ \theta = \arg z

Modulus-argument and exponential forms

The same number in three equivalent forms; $r$ is the distance from the origin and $\theta$ the angle to the positive real axis.

|z_1 z_2| = |z_1||z_2|, \qquad \arg(z_1 z_2) = \arg z_1 + \arg z_2

Multiplication rule

Multiplication multiplies moduli and adds arguments — a rotation combined with an enlargement.

Worked example

Cartesian to modulus-argument, then multiply

Write $z = 1 + \sqrt{3}\,\mathrm{i}$ in modulus-argument form and hence evaluate $z^4$ , giving the answer in Cartesian form.

Modulus
$|z| = \sqrt{1^2 + (\sqrt3)^2} = \sqrt{1+3} = 2$ .
Argument
$z$ is in the first quadrant with $\tan\theta = \sqrt3/1 = \sqrt3$ , so $\theta = \tfrac{\pi}{3}$ . Hence $z = 2\operatorname{cis}\tfrac{\pi}{3}$ .
Raise to the 4th power
By de Moivre, $z^4 = 2^4\operatorname{cis}\tfrac{4\pi}{3} = 16\operatorname{cis}\tfrac{4\pi}{3}$ .
$z^4 = 16\left(\cos\tfrac{4\pi}{3} + \mathrm{i}\sin\tfrac{4\pi}{3}\right)$
Convert back
$\cos\tfrac{4\pi}{3} = -\tfrac12$ , $\sin\tfrac{4\pi}{3} = -\tfrac{\sqrt3}{2}$ , so $z^4 = 16\left(-\tfrac12 - \tfrac{\sqrt3}{2}\mathrm{i}\right) = -8 - 8\sqrt3\,\mathrm{i}$ .

Result: $z = 2\operatorname{cis}\tfrac{\pi}{3}$ and $z^4 = -8 - 8\sqrt3\,\mathrm{i}$ .

Typical mistakes

Taking $\arg z = \arctan(b/a)$ without adjusting for the quadrant, giving an argument that is out by $\pi$ .
Giving an argument outside the principal range $(-\pi, \pi]$ when the principal value is required.

Active revision

Express $z = -\sqrt{3} + \mathrm{i}$ in the form $r(\cos\theta + \mathrm{i}\sin\theta)$ with $\theta$ in the principal range, and hence write down $z^2$ in modulus-argument form.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

de Moivre's theorem and its applications#

AdvancedAQA 7367 §3.2DfE Further Mathematics — Complex numbers

Powers of z rotate around the unit circle

Fig. 3With

z = \operatorname{cis}30^{\circ}

, each successive power rotates a further

30^{\circ}

and stays on the unit circle.

de Moivre's theorem states that

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

for every integer

n

. In exponential form it is nothing more than the index law

(\mathrm{e}^{\mathrm{i}\theta})^n = \mathrm{e}^{\mathrm{i}n\theta}

, but the trigonometric statement is where its power lies: raising to a power multiplies the argument by

n

while raising the modulus to the

n

th power.

For a positive integer

n

the theorem is proved by induction. The base case

n=1

is immediate. Assuming

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

, multiply both sides by

(\cos\theta+\mathrm{i}\sin\theta)

and expand; collecting real and imaginary parts and using the compound-angle formulae gives

\cos(k+1)\theta + \mathrm{i}\sin(k+1)\theta

, completing the step. The result then extends to negative integers using

z^{-1} = \bar z/|z|^2

and to

n=0

trivially.

A principal application is deriving multiple-angle identities. Expanding

(\cos\theta + \mathrm{i}\sin\theta)^n

by the binomial theorem and equating it to

\cos n\theta + \mathrm{i}\sin n\theta

lets you read off

\cos n\theta

from the real part and

\sin n\theta

from the imaginary part. For instance

\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta = 4\cos^3\theta - 3\cos\theta

, a result that would be tedious to obtain from the compound-angle formulae alone.

The reverse direction is equally useful: writing

z = \mathrm{e}^{\mathrm{i}\theta}

gives

z + z^{-1} = 2\cos\theta

and

z - z^{-1} = 2\mathrm{i}\sin\theta

, and more generally

z^n + z^{-n} = 2\cos n\theta

. These let you express powers such as

\cos^4\theta

\sin^5\theta

as sums of multiple angles — exactly the form needed to integrate them. de Moivre therefore connects the algebra of complex numbers to the whole of trigonometry and, through it, to integration.

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

de Moivre's theorem

Raising to the power $n$ multiplies the argument by $n$ ; valid for all integers $n$ .

z^n + \frac{1}{z^n} = 2\cos n\theta, \qquad z^n - \frac{1}{z^n} = 2\mathrm{i}\sin n\theta \quad (z = \mathrm{e}^{\mathrm{i}\theta})

Powers to multiple angles

The tool for linearising powers of sine and cosine before integrating.

Worked example

Deriving cos 3θ

Use de Moivre's theorem to express $\cos 3\theta$ in terms of $\cos\theta$ .

Apply de Moivre
$\cos 3\theta + \mathrm{i}\sin 3\theta = (\cos\theta + \mathrm{i}\sin\theta)^3$ .
Binomial expansion
$(\cos\theta + \mathrm{i}\sin\theta)^3 = \cos^3\theta + 3\cos^2\theta(\mathrm{i}\sin\theta) + 3\cos\theta(\mathrm{i}\sin\theta)^2 + (\mathrm{i}\sin\theta)^3$ .
$= (\cos^3\theta - 3\cos\theta\sin^2\theta) + \mathrm{i}(3\cos^2\theta\sin\theta - \sin^3\theta)$
Equate real parts
$\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta$ .
Eliminate sin²θ
Using $\sin^2\theta = 1 - \cos^2\theta$ : $\cos 3\theta = \cos^3\theta - 3\cos\theta(1-\cos^2\theta) = 4\cos^3\theta - 3\cos\theta$ .

Result: $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ .

Typical mistakes

Sign errors when expanding powers of $\mathrm{i}$ (recall $\mathrm{i}^2=-1$ , $\mathrm{i}^3=-\mathrm{i}$ , $\mathrm{i}^4=1$ ) in the binomial expansion.
Forgetting that $\sin^2\theta = 1 - \cos^2\theta$ must be used to write $\cos n\theta$ purely in terms of $\cos\theta$ .

Active revision

Use de Moivre's theorem to show that $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ .

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

Roots of complex numbers and roots of unity#

AdvancedAQA 7367 §3.2DfE Further Mathematics — Complex numbers

The cube roots of unity

Fig. 4The three cube roots of unity sit at the vertices of an equilateral triangle on the unit circle, one at

z=1

Every non-zero complex number has exactly

n

distinct

n

th roots. To find them, write the number in modulus-argument form

w = r\operatorname{cis}\theta

and add whole multiples of

2\pi

to the argument before taking the root: the

n

th roots are

r^{1/n}\operatorname{cis}\dfrac{\theta + 2\pi k}{n}

for

k = 0, 1, \dots, n-1

. Because the argument increases by

2\pi/n

from one root to the next, the roots have equal modulus

r^{1/n}

and equally spaced arguments.

Geometrically this means the

n

th roots lie at the vertices of a regular

n

-gon inscribed in a circle of radius

r^{1/n}

centred at the origin. Once you have found one root, the others are obtained simply by rotating it repeatedly through

2\pi/n

. This symmetry is both a computational shortcut and a check: if your roots are not evenly spaced with equal modulus, something has gone wrong.

The

n

th roots of unity are the special case

w = 1

: they are

1, \omega, \omega^2, \dots, \omega^{n-1}

where

\omega = \operatorname{cis}\dfrac{2\pi}{n}

. They form a regular

n

-gon on the unit circle with one vertex at

1

. A key algebraic property is that they sum to zero for

n \geq 2

— since they are the roots of

z^n - 1 = 0

, the sum of the roots equals minus the coefficient of

z^{n-1}

, which is

0

— a fact with a pleasing geometric reading as balanced vectors cancelling.

The roots of unity also form a multiplicative structure: multiplying any root of a number

w

by a root of unity gives another root of

w

, which is why finding all roots reduces to finding one and multiplying by

1, \omega, \dots, \omega^{n-1}

. For real coefficients the non-real roots still pair as conjugates, consistent with the

n

-gon being symmetric in the real axis. This blend of algebra (roots of a polynomial) and geometry (a regular polygon) is one of the most elegant results in the whole A-Level.

z_k = r^{1/n}\operatorname{cis}\!\left(\frac{\theta + 2\pi k}{n}\right), \quad k = 0,1,\dots,n-1

The n nth roots of r cis θ

Equal modulus $r^{1/n}$ , arguments spaced by $2\pi/n$ : the vertices of a regular $n$ -gon.

1 + \omega + \omega^2 + \cdots + \omega^{n-1} = 0, \qquad \omega = \operatorname{cis}\tfrac{2\pi}{n}

Sum of the n roots of unity

The roots of $z^n-1=0$ sum to zero for $n\geq 2$ ; the balanced vectors cancel.

Worked example

The cube roots of 8i

Solve $z^3 = 8\mathrm{i}$ , giving each root in modulus-argument form.

Write 8i in mod-arg form
$8\mathrm{i}$ has modulus $8$ and argument $\tfrac{\pi}{2}$ , so $8\mathrm{i} = 8\operatorname{cis}\!\left(\tfrac{\pi}{2} + 2\pi k\right)$ .
Take cube roots
$z = 8^{1/3}\operatorname{cis}\!\left(\dfrac{\pi/2 + 2\pi k}{3}\right) = 2\operatorname{cis}\!\left(\dfrac{\pi}{6} + \dfrac{2\pi k}{3}\right)$ for $k=0,1,2$ .
$z = 2\operatorname{cis}\!\left(\tfrac{\pi}{6} + \tfrac{2\pi k}{3}\right)$
List the roots
$k=0$ : $2\operatorname{cis}\tfrac{\pi}{6}$ ; $k=1$ : $2\operatorname{cis}\tfrac{5\pi}{6}$ ; $k=2$ : $2\operatorname{cis}\tfrac{3\pi}{2} = 2\operatorname{cis}\!\left(-\tfrac{\pi}{2}\right)$ .
Interpret
In Cartesian form these are $\sqrt3 + \mathrm{i}$ , $-\sqrt3 + \mathrm{i}$ and $-2\mathrm{i}$ — the vertices of an equilateral triangle of circumradius $2$ centred at the origin.

Result: $z = 2\operatorname{cis}\tfrac{\pi}{6},\ 2\operatorname{cis}\tfrac{5\pi}{6},\ 2\operatorname{cis}\!\left(-\tfrac{\pi}{2}\right)$ ; equally $\sqrt3+\mathrm{i},\ -\sqrt3+\mathrm{i},\ -2\mathrm{i}$ .

Typical mistakes

Producing fewer than $n$ roots, or repeating a root, by not letting $k$ run over a full set $0,1,\dots,n-1$ .
Leaving an argument outside the principal range, or forgetting to take the $n$ th root of the modulus.

Active revision

Solve $z^3 = 8\mathrm{i}$ , giving the three roots in the form $r\operatorname{cis}\theta$ with $\theta$ in the principal range, and describe their geometric arrangement.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Further Mathematics 7367 specification (AQA)

Loci and regions in the Argand diagram#

AdvancedAQA 7367 §3.2DfE Further Mathematics — Complex numbers

A circle locus and a half-line locus

Fig. 5The circle is

|z-(2+\mathrm{i})|=2

, centre

2+\mathrm{i}

; the ray is

\arg(z-1)=\tfrac{\pi}{4}

, starting at

1

(excluded).

A locus in the Argand diagram is the set of points

z

satisfying a condition, and the key to sketching it is to read every modulus as a distance and every argument as a direction. Because

|z - a|

is the distance from

z

to the fixed point

a

, the equation

|z - a| = r

describes all points a fixed distance

r

from

a

— a circle of radius

r

centred at

a

. The corresponding inequality

|z - a| \leq r

shades the closed disc.

The condition

|z - a| = |z - b|

says

z

is equidistant from the two fixed points

a

and

b

, so the locus is the perpendicular bisector of the line segment joining

a

and

b

. This is a straight line, and it is best found geometrically (midpoint and perpendicular direction) rather than by grinding through the algebra, though squaring

|z-a|^2 = |z-b|^2

also yields the line if needed.

An argument condition

\arg(z - a) = \alpha

fixes the direction from

a

z

, so the locus is a half-line (a ray) starting at

a

— but not including

a

itself, since

\arg 0

is undefined — and making angle

\alpha

with the positive real direction. Only the ray in the specified direction is included; the opposite ray, with argument

\alpha + \pi

, is not. Marking the excluded endpoint with an open circle and drawing only the correct ray are both examined.

Harder problems combine conditions or ask for the point on a locus that is nearest to, or furthest from, the origin. Here the geometry pays off: the nearest point of a circle to the origin lies on the line from the origin through the centre, at distance

|\,\text{centre}\,| - r

, and the greatest value of

|z|

on the circle is

|\,\text{centre}\,| + r

. Solving such a problem by sketching the circle and the relevant line is far quicker and more reliable than a purely algebraic attack.

|z - a| = r \ \text{(circle, centre } a,\ \text{radius } r), \qquad |z-a| = |z-b| \ \text{(perpendicular bisector of } a,b)

Distance loci

Circle and perpendicular bisector — the two loci that come from modulus conditions.

\arg(z - a) = \alpha \ \text{(half-line from } a,\ \text{excluding } a,\ \text{at angle } \alpha)

Argument locus

A ray, not a full line: only the direction $\alpha$ from $a$ is included.

Worked example

Greatest and least modulus on a circle

The complex number $z$ satisfies $|z - (3 + 4\mathrm{i})| = 2$ . Find the greatest and least possible values of $|z|$ .

Identify the circle
The locus is a circle of radius $2$ centred at the point $c = 3 + 4\mathrm{i}$ .
Distance of the centre from O
$|c| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ .
Use the line through O and the centre
The greatest and least distances from the origin to a point on the circle are $|c| + r$ and $|c| - r$ .
$|z|_{\max} = |c| + r = 5 + 2 = 7, \qquad |z|_{\min} = |c| - r = 5 - 2 = 3$
State the answer
So $|z|$ ranges between $3$ and $7$ , both achieved where the line $Oc$ meets the circle.

Result: Greatest $|z| = 7$ , least $|z| = 3$ .

Typical mistakes

Drawing the whole line instead of a half-line for an $\arg(z-a)=\alpha$ condition, or including the excluded endpoint.
Misreading $|z - a| = r$ as a circle centred at the origin rather than at $a$ , or taking the centre as $-a$ with the wrong sign.

Active revision

Sketch on one Argand diagram the loci $|z - (2 + \mathrm{i})| = 2$ and $\arg(z - 1) = \tfrac{\pi}{4}$ , and describe each in words.

Active recall

Recall the key points — then reveal.

Sources: Further mathematics: AS and A level content (GCE subject content) (Department for Education)

References & sources

Sources

Department for Education

Further mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Further Mathematics 7367 specification

Complex numbers

Key points

Division and a conjugate root

Key points

Cartesian to modulus-argument, then multiply

Key points

Deriving cos 3θ

Key points

The cube roots of 8i

Key points

Greatest and least modulus on a circle

Sources