Complex Numbers

CIE A-Level Mathematics· 20 min read

1. Definition and Basic Representation★★☆☆☆⏱ 5 min

📘 Definition

Complex Number

z = x + iy

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

Example:

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

2. Arithmetic of Complex Numbers★★★☆☆⏱ 7 min

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 $i^{2} = - 1$ .

📘 Definition

Complex Conjugate

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

For $z = x + i y$ , the conjugate is $z^{*} = x - i y$ , found by flipping the sign of the imaginary part. A key property is $z z^{*} = ∣ z ∣^{2}$ , a real number.

Example:

Conjugate of $5 + 2 i$ is $5 - 2 i$ .

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

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

3. Square Roots of a Complex Number★★★☆☆⏱ 6 min

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

📘 Definition

Square Roots by Comparing Parts

(x+iy)^2 = a + ib

Expanding gives $(x^{2} - y^{2}) + 2 x y i = a + ib$ . Comparing parts: $x^{2} - y^{2} = a$ and $2 x y = b$ . Solve this pair for real $x, y$ ; the two solutions are negatives of each other.

Example:

For $5 + 12 i$ : $x^{2} - y^{2} = 5$ and $2 x y = 12$ .

📐 Worked Example

Find the square roots of $5 + 12 i$ , giving each in the form $x + i y$ .

1
Let $5 + 12 i = x + i y$ 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 $3 + 2 i$ and $- 3 - 2 i$ , i.e. $\pm (3 + 2 i)$ .

Exam tip:

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

4. Modulus and Principal Argument★★★★☆⏱ 6 min

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 $- π < ar g (z) \leq π$ for CIE exams.

📘 Definition

Modulus and Argument

$∣ z ∣$ (modulus), $ar g (z)$ (argument)

For $z = x + i y$ : $∣ z ∣ = x^{2} + y^{2}$ , and $tan (ar g (z)) = \frac{y}{x}$ . Always adjust the angle to match the quadrant of the complex number.

📐 Worked Example

Find the modulus and principal argument of $z = - 1 + i 3$ . Give your answer in radians.

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

Exam tip:

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

5. Modulus-Argument (Polar) Form★★★★☆⏱ 7 min

Instead of $x + i y$ , a complex number can be written using its modulus $r = ∣ z ∣$ and argument $θ = ar g (z)$ . This modulus-argument (polar) form makes multiplication and division especially simple.

📘 Definition

Modulus-Argument (Polar) and Exponential Form

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

Here $r = ∣ z ∣ = x^{2} + y^{2} \geq 0$ and $θ = ar g (z)$ taken in the principal range $- π < θ \leq π$ . The exponential form $r e^{i θ}$ is an equivalent notation for the same number, with $x = r cos θ$ and $y = r sin θ$ .

Example:

$- 1 + i 3 = 2 (cos \frac{2 π}{3} + i sin \frac{2 π}{3}) = 2 e^{i 2 π /3}$ .

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.

📘 Definition

Multiplication and Division in Polar Form

$∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣∣ z_{2} ∣$ and $ar g (z_{1} z_{2}) = ar g z_{1} + ar g z_{2}$ ; $\frac{z _{1}}{z _{2}} = \frac{∣ z _{1} ∣}{∣ z _{2} ∣}$ and $ar g (\frac{z _{1}}{z _{2}}) = ar g z_{1} - ar g z_{2}$ . After adding or subtracting, adjust the result by $\pm 2 π$ so it lies in $- π < θ \leq π$ .

Example:

If $ar g z_{1} = \frac{3 π}{4}$ and $ar g z_{2} = \frac{3 π}{4}$ , then $ar g (z_{1} z_{2}) = \frac{3 π}{2}$ , adjusted to $\frac{3 π}{2} - 2 π = - \frac{π}{2}$ .

📐 Worked Example

Let $z_{1} = 2 (cos \frac{π}{3} + i sin \frac{π}{3})$ and $z_{2} = 3 (cos \frac{π}{4} + i sin \frac{π}{4})$ . 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 $\frac{7 π}{12}$ and $\frac{π}{12}$ already lie in $- π < θ \leq π$ , so no adjustment is needed.

Exam tip:

After adding or subtracting arguments, always bring the result back into the principal range $- π < θ \leq π$ before stating the final argument.

6. Geometric Effects on the Argand Diagram★★★☆☆⏱ 5 min

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

📘 Definition

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 $ar g z_{2}$ about the origin.

Example:

Multiplying by $i$ (modulus 1, argument $\frac{π}{2}$ ) rotates a point $9 0^{\circ}$ 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) $i z$ , 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 $\frac{π}{2}$ (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 $9 0^{\circ}$ anticlockwise turn about the origin.

7. Loci in the Argand Diagram★★★★☆⏱ 7 min

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 $z = x + i y$ and interpret $∣ z - a ∣$ as the distance from $z$ to the fixed point $a$ .

📘 Definition

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: $ar g (z - a) = α$ is a ray (half-line) starting at $a$ — with the endpoint $a$ excluded — making angle $α$ with the positive real direction.

Example:

$∣ z - (1 + 2 i) ∣ = 3$ 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) $ar g (z - 1) = \frac{π}{4}$ .

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 $\frac{π}{4}$ . This is a half-line starting at the point $(1, 0)$ (open endpoint) going up and to the right at $4 5^{\circ}$ 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.

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.

8. Complex Conjugate Root Theorem★★★☆☆⏱ 5 min

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.

📘 Definition

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

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

9. Common Pitfalls

Wrong move:

Claiming $i^{2} = 1$ by using $i = - 1$ so $i^{2} = (- 1) (- 1) = 1 = 1$

Why:

The rule $a b = ab$ only applies for non-negative real $a$ and $b$ , so this extension is invalid.

Correct move:

Always use the definition $i^{2} = - 1$ directly, do not extend real-number root rules to negative numbers.

Wrong move:

Writing $ar g (z) = arctan (\frac{y}{x})$ directly without adjusting for quadrant

Why:

$arctan$ only returns values between $- \frac{π}{2}$ and $\frac{π}{2}$ , so it will give the wrong angle for complex numbers in the second and third quadrants.

Correct move:

Always check the quadrant of your complex number, and add or subtract $π$ as needed to get the principal argument.

Wrong move:

When dividing complex numbers, multiply only the numerator by the conjugate of the denominator

Why:

This changes the value of the fraction, leading to an incorrect result.

Correct move:

Always multiply both the numerator and denominator by the conjugate to keep the fraction equivalent.

Wrong move:

Using the conjugate root theorem for polynomials with complex coefficients

Why:

The theorem only holds when all coefficients of the polynomial are real.

Correct move:

Only use the conjugate root theorem if the question explicitly states the polynomial has real coefficients.

Wrong move:

Giving only one square root of a complex number, or keeping a solution where $x^{2}$ comes out negative.

Why:

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 move:

Solve $x^{2} - y^{2} = a$ and $2 x y = b$ for real $x, y$ , reject non-real solutions, and give both roots $\pm (x + i y)$ .

Wrong move:

Leaving $ar g (z_{1} z_{2})$ or $ar g (z_{1} / z_{2})$ outside $- π < θ \leq π$ after adding or subtracting arguments.

Why:

The sum or difference of two principal arguments can fall outside the principal range, so it is no longer the principal argument.

Correct move:

After adding or subtracting, add or subtract $2 π$ as needed to bring the argument back into $- π < θ \leq π$ .

Wrong move:

Sketching $ar g (z - a) = α$ as a full straight line through $a$ , or including the point $a$ itself.

Why:

The argument fixes a single direction from $a$ , so only one ray qualifies; and $ar g 0$ is undefined, so $a$ is excluded.

Correct move:

Draw a single half-line from $a$ at angle $α$ , mark the endpoint $a$ with an open circle, and do not draw the opposite direction $α + π$ .

10. Quick Reference Cheatsheet

Property	Rule for $z = x + i y, z_{1} = x_{1} + i y_{1}, z_{2} = x_{2} + i y_{2}$
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	$z_{1} z_{2} = (x_{1} x_{2} - y_{1} y_{2}) + i (x_{1} y_{2} + x_{2} y_{1})$
Division	$\frac{z _{1}}{z _{2}} = \frac{z _{1} z _{2}^{*}}{∣ z _{2} ∣ ^{2}}$
Conjugate	$z^{} = x - i y, z z^{} = ∣ z ∣^{2}$ (always real)
Modulus	$∣ z ∣ = x^{2} + y^{2}$
Argument	$tan ar g (z) = \frac{y}{x}$ , adjust for quadrant, $- π < ar g (z) \leq π$
Polar form	$z = r (cos θ + i sin θ) = r e^{i θ}$ , $r = ∣ z ∣$ , $θ = ar g (z)$
Polar multiply	$∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣∣ z_{2} ∣$ , $ar g (z_{1} z_{2}) = ar g z_{1} + ar g z_{2}$
Polar divide	$\frac{z _{1}}{z _{2}} = \frac{∣ z _{1} ∣}{∣ z _{2} ∣}$ , $ar g (\frac{z _{1}}{z _{2}}) = ar g z_{1} - ar g z_{2}$
Square roots	$a + ib = x + i y$ : solve $x^{2} - y^{2} = a$ , $2 x y = b$ ; roots $\pm (x + i y)$
Locus: circle	$∣ z - a ∣ = r$ : circle, centre $a$ , radius $r$
Locus: bisector	$∣ z - a ∣ = ∣ z - b ∣$ : perpendicular bisector of $a$ , $b$
Locus: half-line	$ar g (z - a) = α$ : half-line from $a$ (excl.), angle $α$
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.