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Complex Numbers

Lessons

Senior 6 · Mathematics

Complex Numbers

5 lessons

Lesson 1 of 5

The imaginary unit, the Argand diagram and the modulus

25 min

Where the real numbers run out

In S4 (Linear and Quadratic Equations) you solved $a x^{2} + b x + c = 0$ with $\Delta = b^2 - 4ac$ , and when $\Delta \lt 0$ you wrote "no real roots". Take $x^{2} + 4 x + 13 = 0$ : here $\Delta = 16 - 52 = -36$ , and the formula asks for $\sqrt{-36}$ . No real number squares to $- 36$ , because the square of every real number is at least 0. This unit builds a larger set of numbers, the complex numbers, in which every quadratic equation (and, as you will see, every polynomial equation) has roots. Engineers use the same numbers every day to analyse alternating current, which Lesson 5 shows.

Definition

Imaginary unit: the number $i$ whose square is $- 1$ :

i^{2} = - 1.

Square root of a negative real number: for a real number $a \gt 0$ , $\sqrt{-a} = i\sqrt{a}$ , because $(i\sqrt{a})^2 = i^2 a = -a$ .

Example: $\sqrt{-36} = 6i$ and $\sqrt{-7} = i\sqrt{7}$ . Electrical engineers write $j$ instead of $i$ , because $i$ already stands for current.

Warning

The rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ is only for $a, b \ge 0$ . Using it on negative numbers gives nonsense: $\sqrt{-4}\,\sqrt{-9} = (2i)(3i) = 6i^2 = -6$ , not $\sqrt{36} = 6$ . Always convert $\sqrt{-a}$ to $i\sqrt{a}$ first, then multiply.

Powers of $i$ repeat every four

Multiply by $i$ again and again, and use $i^{2} = - 1$ at each step:

$n$	0	1	2	3	4	5	6	7
$i^{n}$	$1$	$i$	$- 1$	$- i$	$1$	$i$	$- 1$	$- i$

Since $i^{4} = (i^{2})^{2} = (- 1)^{2} = 1$ , the pattern repeats with period 4. If $n = 4 q + r$ with $r \in \{0, 1, 2, 3\}$ , then

i^n = (i^4)^q \, i^r = 1^q \, i^r = i^r .

So only the remainder of $n$ on division by 4 matters. It follows that $i + i^{2} + i^{3} + i^{4} = i - 1 - i + 1 = 0$ : any four consecutive powers of $i$ add up to 0.

Worked example 1: powers of $i$

Simplify (a) $i^{2027}$ (b) $i^{- 3}$ (c) $i + i^2 + i^3 + \cdots + i^{10}$ .

Solution

$2027 = 4 \times 506 + 3$ , so $i^{2027} = i^3 = -i$ .
First, $\frac{1}{i} = \frac{i}{i^2} = -i$ . Then $i^{-3} = \frac{1}{i^3} = \frac{1}{-i} = -\frac{1}{i} = -(-i) = i$ . Check with the rule: $-3 = 4 \times (-1) + 1$ , so $i^{-3} = i^1 = i$ . ✓
The first eight terms form two blocks of four consecutive powers, each adding to 0. What is left is $i^9 + i^{10} = i + (-1) = -1 + i$ .

Definition

Complex number: a number of the form $z = a + b i$ , where $a$ and $b$ are real numbers and $i^{2} = - 1$ . This is the algebraic (Cartesian) form of $z$ .

Notation: $a = \operatorname{Re}(z)$ is the real part and $b = \operatorname{Im}(z)$ is the imaginary part. Note that the imaginary part is the real number $b$ , not $b i$ . The set of all complex numbers is $\mathbb{C}$ .

If $b = 0$ , $z = a$ is purely real; if $a = 0$ and $b \ne 0$ , $z = b i$ is purely imaginary. So $\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}$ .

Example: $z = -5 + \frac{3}{4}i$ has $\operatorname{Re}(z) = -5$ and $\operatorname{Im}(z) = \frac{3}{4}$ .

Definition

Equality of complex numbers: $a + b i = c + d i$ (with $a, b, c, d$ real) if and only if $a = c$ and $b = d$ .

Why: suppose $a + b i = c + d i$ but $b \ne d$ . Then $i = \frac{a - c}{d - b}$ would be a real number, and its square would be at least 0, not $- 1$ . So $b = d$ , and then $a = c$ . One equation between complex numbers therefore gives two real equations.

Worked example 2: two equations from one

Find the real numbers $x$ and $y$ such that $(2 x - y) + (x + 3 y) i = 5 - i$ .

Solution Equate real parts and imaginary parts:

2x - y = 5 \quad\text{and}\quad x + 3y = -1.

From the first, $y = 2 x - 5$ . Substituting: $x + 3 (2 x - 5) = - 1$ , so $7 x = 14$ , $x = 2$ and $y = - 1$ .

Check: $2 (2) - (- 1) = 5$ ✓ and $2 + 3 (- 1) = - 1$ ✓.

Adding and subtracting

Treat $i$ like a letter in algebra and collect the real parts and the imaginary parts:

(a + b i) + (c + d i) = (a + c) + (b + d) i,

(a + b i) - (c + d i) = (a - c) + (b - d) i .

For example, $(4 - 3 i) + (- 1 + 5 i) = 3 + 2 i$ and $(4 - 3 i) - (- 1 + 5 i) = 5 - 8 i$ . Because the real parts and the imaginary parts are added as real numbers, addition in $\mathbb{C}$ is commutative and associative, $0 = 0 + 0 i$ is the identity, and $- z = - a - b i$ is the opposite of $z$ .

Definition

Argand diagram (complex plane): a plane with a horizontal real axis and a vertical imaginary axis, in which the complex number $z = a + b i$ is drawn as the point $(a, b)$ , or as the vector $\overrightarrow{OZ} = \begin{pmatrix} a \\ b \end{pmatrix}$ from the origin to that point.

The point that represents $z$ is called the image of $z$ , and $z$ is called the affix of the point.

Each complex number $a + b i$ is the point $(a, b)$ . Purely real numbers lie on the real axis; purely imaginary numbers lie on the imaginary axis.

Because $a + b i$ behaves like the vector $\begin{pmatrix} a \\ b \end{pmatrix}$ , adding complex numbers is adding vectors, which you met in S4 (Vector Space of Real Numbers). The sum is found with the parallelogram rule, and $z_{2} - z_{1}$ is the vector that goes from the point $z_{1}$ to the point $z_{2}$ .

Adding complex numbers adds their vectors: $z_{1} + z_{2}$ is the diagonal of the parallelogram built on $z_{1}$ and $z_{2}$ .

Definition

Modulus of $z = a + b i$ : the distance from the origin to the point $z$ ,

|z| = \sqrt{a^2 + b^2} \ge 0 .

Distance between two points: the distance between the points $z_{1}$ and $z_{2}$ is $∣ z_{1} - z_{2} ∣$ , since $z_{1} - z_{2}$ is the vector joining them.

Example: $|3 + 4i| = \sqrt{9 + 16} = 5$ . For a real number, the modulus is the absolute value: $∣ - 7 ∣ = 7$ .

By Pythagoras, $|3 + 4i| = \sqrt{3^2 + 4^2} = 5$ : the modulus is the distance from O to the point.

Worked example 3: modulus and distance

(a) Find $∣ 5 - 12 i ∣$ and $∣ - 2 + i ∣$ . (b) Find the distance between the points $1 + 2 i$ and $4 - 2 i$ .

Solution

$|5 - 12i| = \sqrt{5^2 + (-12)^2} = \sqrt{169} = 13$ , and $|-2 + i| = \sqrt{4 + 1} = \sqrt{5} \approx 2.24$ .
$(1 + 2 i) - (4 - 2 i) = - 3 + 4 i$ , so the distance is $|-3 + 4i| = \sqrt{9 + 16} = 5$ .

Loci defined by distances

A locus is the set of all points that satisfy a condition. Read $∣ z - z_{0} ∣$ as "the distance from $z$ to $z_{0}$ ", and many loci can be drawn at sight.

$∣ z - z_{0} ∣ = r$ (with $r \gt 0$ ): all points at distance $r$ from $z_{0}$ , a circle with centre $z_{0}$ and radius $r$ . The inequality $|z - z_0| \le r$ is the disc, and $|z - z_0| \gt r$ is the outside of the circle.
$∣ z - a ∣ = ∣ z - b ∣$ (with $a \ne b$ ): all points equally far from $a$ and $b$ , the perpendicular bisector of the segment joining them.

To find the Cartesian equation, put $z = x + y i$ and square both moduli.

A circle. The condition $∣ z + 1 - 2 i ∣ = 3$ is $∣ z - (- 1 + 2 i) ∣ = 3$ . With $z = x + y i$ : $∣ (x + 1) + (y - 2) i ∣ = 3$ , so

(x + 1)^{2} + (y - 2)^{2} = 9,

the circle with centre $(- 1, 2)$ and radius 3.

$∣ z - (- 1 + 2 i) ∣ = 3$ : every point on the circle is 3 units from $- 1 + 2 i$ . The shaded disc is $|z + 1 - 2i| \le 3$ .

Worked example 4: a perpendicular bisector

Find the Cartesian equation of the locus $∣ z - 2 ∣ = ∣ z - 4 i ∣$ , and describe it.

Solution Put $z = x + y i$ . Then $∣ (x - 2) + y i ∣ = ∣ x + (y - 4) i ∣$ . Square both sides:

(x - 2)^{2} + y^{2} = x^{2} + (y - 4)^{2}

x^{2} - 4 x + 4 + y^{2} = x^{2} + y^{2} - 8 y + 16

-4x + 8y - 12 = 0 \quad\Longrightarrow\quad x - 2y + 3 = 0, \text{ i.e. } y = \frac{x + 3}{2}.

Check: the midpoint of $2$ and $4 i$ is $(1, 2)$ , and $1 - 4 + 3 = 0$ ✓. The segment from $(2, 0)$ to $(0, 4)$ has gradient $- 2$ , and the locus has gradient $\frac{1}{2}$ ; their product is $- 1$ , so they are perpendicular ✓.

Every point of the blue line is equally far from $2$ and from $4 i$ . Its Cartesian equation is $x - 2 y + 3 = 0$ .

Important

Condition on $z$	Locus in the Argand diagram
$∣ z - z_{0} ∣ = r$	circle, centre $z_{0}$ , radius $r$
$\|z - z_0\| \lt r$	inside of that circle (open disc)
$∣ z - a ∣ = ∣ z - b ∣$	perpendicular bisector of the segment from $a$ to $b$
$\|z - a\| \lt \|z - b\|$	the half-plane on the side of $a$

Warning

Watch the signs of the centre. In $∣ z + 1 - 2 i ∣ = 3$ the centre is $- 1 + 2 i$ , not $1 - 2 i$ . Always rewrite the condition as $|z - (\text{centre})|$ before reading off the centre.

Tip

In an exam, sketch the locus first and write the Cartesian equation second. The sketch tells you what shape to expect (circle or straight line), so an algebra slip shows up at once.

Welch Labs: Imaginary Numbers Are Real, Part 1 (about 5 min). Watch for how a graph shows the real roots of a polynomial, and why an equation with no real roots pushed mathematicians towards a second, imaginary axis.

✦

Practice

Try each one before opening the answers. The last two are harder.

Simplify: (a) $\sqrt{-49}$ (b) $i^{15}$ (c) $i^{- 2}$ (d) $(2 - 5 i) - (- 3 + i)$ .
Find the real numbers $x$ and $y$ such that $(x + 2 y) + (3 x - y) i = 4 + 5 i$ .
Plot $z_{1} = 2 - 3 i$ , $z_{2} = - 4$ and $z_{3} = 3 i$ on one Argand diagram, and add the point $z_{1} + z_{2}$ .
Find the modulus of (a) $- 6 + 8 i$ (b) $1 - i$ (c) $2 i - 7$ .
Find the distance between the points $3 - i$ and $- 1 + 2 i$ .
Describe the locus $∣ z - 3 + i ∣ = 5$ and give its Cartesian equation.
Find the Cartesian equation of the locus $∣ z + 2 ∣ = ∣ z - 2 i ∣$ , and check it with a sketch.
The point $z$ moves on the circle $∣ z - 3 - 4 i ∣ = 2$ . Find the greatest and the least values of $∣ z ∣$ . Explain your answer with a sketch.

Show answersHide answers

(a) $7 i$ (b) $15 = 4 \times 3 + 3$ , so $i^{15} = i^3 = -i$ (c) $i^{-2} = \frac{1}{i^2} = -1$ (d) $5 - 6 i$ .
$x + 2 y = 4$ and $3 x - y = 5$ . Then $y = 3 x - 5$ , so $x + 6 x - 10 = 4$ , $x = 2$ , $y = 1$ .
$z_{1}$ is the point $(2, - 3)$ , $z_{2}$ is $(- 4, 0)$ on the real axis, $z_{3}$ is $(0, 3)$ on the imaginary axis, and $z_{1} + z_{2} = - 2 - 3 i$ is the point $(- 2, - 3)$ .
(a) $\sqrt{36 + 64} = 10$ (b) $\sqrt{2}$ (c) $|-7 + 2i| = \sqrt{53} \approx 7.28$ .
$(3 - i) - (- 1 + 2 i) = 4 - 3 i$ , and $∣ 4 - 3 i ∣ = 5$ .
$∣ z - (3 - i) ∣ = 5$ : the circle with centre $3 - i$ and radius 5, $(x - 3)^{2} + (y + 1)^{2} = 25$ .
$(x + 2)^{2} + y^{2} = x^{2} + (y - 2)^{2}$ gives $4 x + 4 = - 4 y + 4$ , so $y = - x$ . It is the perpendicular bisector of the segment from $(- 2, 0)$ to $(0, 2)$ : it passes through the midpoint $(- 1, 1)$ and has gradient $- 1$ , perpendicular to the segment's gradient $1$ .
The centre $3 + 4 i$ is at distance $∣ 3 + 4 i ∣ = 5$ from O, and the radius is 2. The line from O through the centre meets the circle at distances $5 - 2 = 3$ and $5 + 2 = 7$ from O. So the least value of $∣ z ∣$ is 3 and the greatest is 7.

Sources and credits

Written by Ganzaa: the explanations, the diagrams, the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S6 Core Mathematics (2026–2027); no REB text is reproduced.
Video: Imaginary Numbers Are Real [Part 1: Introduction], Welch Labs, embedded from YouTube.

1Where the real numbers run out

2Powers of i i i repeat every four

3Adding and subtracting

4Loci defined by distances

Where the real numbers run out

Powers of $i$ repeat every four

Adding and subtracting

Loci defined by distances