1.12 Complex numbers in Cartesian form · Educerie Mathematics: analysis and approaches

This whole subtopic is higher level. Nothing in it is on an SL paper.

Educerie · IB Diploma · Mathematics: analysis and approaches

Topic 1 Number and algebra · 1.12 Complex numbers in Cartesian form

Level: HL only, the whole subtopic. If you are SL, none of this is on your papers.
Themes (key concepts): generalization, representation, systems. Complex numbers generalize the real numbers so that every quadratic has roots, and the complex plane represents each one as a point, which turns arithmetic into geometry and gives a new number system with its own rules.
The question this unit answers: what happens if you allow a number whose square is −1, and how do you calculate with, and draw, the numbers that result?
Where it is examined: Paper 1 (no calculator): multiplying and dividing in the form a + bi, finding a modulus and argument exactly, and solving an equation by equating real and imaginary parts (3 to 7 marks). Paper 2: the GDC's complex mode for heavy arithmetic. Paper 3 and every later complex-numbers question (1.13, 1.14) assume all of this is automatic.

What you must be able to do

You must be able to	Level	What it looks like in the exam
Use i² = −1, write square roots of negative numbers in terms of i, and simplify powers of i	HL only	Solving z² − 6z + 34 = 0 as one step of a question (3 marks)
Write a complex number in Cartesian form a + bi and name its real part and imaginary part	HL only	"Find Re(z) and Im(z)" or "given that Im(z) = 0…"
Add, subtract and multiply complex numbers in Cartesian form	HL only	"Given z = 2 − 3i and w = 4 + i, find zw" (2 marks, Paper 1)
Find the conjugate z* and divide by multiplying top and bottom by the conjugate of the denominator	HL only	"Express z/w in the form a + bi" (3 marks, Paper 1)
Solve equations by equating real parts and imaginary parts, including finding square roots of a complex number	HL only	"Find the complex numbers z such that z² = −24 − 10i" (6 marks)
Find the modulus	z	and the argument arg z, choosing the correct quadrant	HL only	"Find the modulus and argument of −√3 + i" (4 marks); exact on Paper 1
Plot complex numbers on the complex plane (Argand diagram), and interpret z*, −z, z + w and z − w there	HL only	"Sketch z and z* on an Argand diagram";	z − w	as a distance

Before you start

You need the quadratic formula and the discriminant from Topic 2 (SL 2.7): when Δ = b² − 4ac is negative, a quadratic has no real roots, and this subtopic is where that changes. You need to expand brackets confidently and simplify surds such as √48 = 4√3. For the argument you need arctan, radians, and the exact values of sin, cos and tan at π/6, π/4 and π/3 from Topic 3. Vectors (AHL 3.12) help but are not required.

1The idea in one paragraph

No real number squares to −1, so x² + 1 = 0 has no real solution. Mathematicians decided to invent one: a number i with i² = −1, and then to allow every number of the form a + bi, where a and b are real. These are complex numbers. You calculate with them exactly as you would with algebra in the letter i, replacing i² by −1 whenever it appears. Every complex number a + bi can be drawn as the point (a, b) on a plane, the complex plane, and then its size (the modulus) is its distance from the origin and its direction (the argument) is its angle from the positive real axis. Division uses the conjugate a − bi to turn the denominator real. That is the whole subtopic, and it is the foundation for 1.13 and 1.14.

2The number i

Define i as a number with

i² = −1.

Then every negative number has square roots. √(−16) = √16 × √(−1) = 4i, and (4i)² = 16i² = −16, as it should. The same goes for −4i. So the equation x² + 16 = 0 has the two solutions x = ±4i.

With i available, the quadratic formula never fails. For x² − 4x + 13 = 0:

x = [4 ± √(16 − 52)] / 2 = [4 ± √(−36)] / 2

= (4 ± 6i) / 2

= 2 ± 3i

Check one: (2 + 3i)² − 4(2 + 3i) + 13 = (4 + 12i + 9i²) − 8 − 12i + 13 = 4 + 12i − 9 − 8 − 12i + 13 = 0. The roots of a quadratic with real coefficients and negative discriminant always come as a pair a ± bi like this. Subtopic 1.14 explains why, and extends it to cubics and quartics.

Powers of i repeat every four: i¹ = i, i² = −1, i³ = i² × i = −i, i⁴ = (i²)² = 1, and then i⁵ = i again. To simplify a large power, divide the exponent by 4 and keep the remainder: i²⁰²⁷ = i^(4 × 506 + 3) = (i⁴)⁵⁰⁶ × i³ = −i. Figure 4, in section 7, shows why the cycle has length four.

About the name. Descartes used "imaginary" as a put-down, and the name stuck. The guide asks a fair TOK question: would students find these numbers easier if they had a less off-putting name? Complex numbers are exactly as real as negative numbers, which were once resisted for the same reasons. Electrical engineers use them every day: the impedance of a circuit combines resistance and reactance into one number R + Xi (engineers write j instead of i, because i already means current).

3Cartesian form: real part and imaginary part

A complex number written as

z = a + bi, with a and b real,

is in Cartesian form. The number a is the real part, written Re(z), and b is the imaginary part, Im(z). For z = 5 − 2i, Re(z) = 5 and Im(z) = −2. The imaginary part is the real number b, not bi. That is a common slip and it costs marks.

When b = 0, z is an ordinary real number: the reals sit inside the complex numbers. When a = 0 and b ≠ 0, z is called purely imaginary, such as 3i.

Equal complex numbers. Two complex numbers are equal only if their real parts are equal and their imaginary parts are equal.

a + bi = c + di exactly when a = c and b = d.

One equation between complex numbers is therefore two equations between real numbers. This is the most useful fact in the subtopic, and section 6 is built on it.

4Adding, subtracting and multiplying

Treat i as a letter, collect real parts with real parts and imaginary parts with imaginary parts, and replace i² by −1. Take z = 3 + 2i and w = 1 + i.

z + w = (3 + 1) + (2 + 1)i = 4 + 3i

z − w = (3 − 1) + (2 − 1)i = 2 + i

zw = (3 + 2i)(1 + i) = 3 + 3i + 2i + 2i²

= 3 + 5i − 2 = 1 + 5i

z² = (3 + 2i)² = 9 + 12i + 4i² = 5 + 12i

Multiplication is just expanding two brackets. The one new step is the last, 2i² = −2, which moves a term from the imaginary part into the real part. Forgetting that step is the commonest error in the whole topic.

5The conjugate, and division

The complex conjugate of z = a + bi is

z* = a − bi: the same real part, the opposite imaginary part.

Multiplying a complex number by its conjugate always gives a real number, because the i terms cancel:

z z* = (a + bi)(a − bi) = a² − abi + abi − b² i² = a² + b²

That is what makes division work. To divide by a complex number, multiply the top and the bottom by the conjugate of the bottom. The denominator becomes real, and you can split the answer into a real part and an imaginary part. It is the same idea as rationalising the denominator of 1/(2 + √3) with surds.

Worked example 1. Paper 1. With z = 3 + 2i and w = 1 + i, find z/w in the form a + bi.

z / w = (3 + 2i)/(1 + i) × (1 − i)/(1 − i)

= (3 − 3i + 2i − 2i²) / (1 − i²)

= (3 − i + 2) / (1 + 1)

= (5 − i)/2 = 5/2 − (1/2)i

Check by multiplying back: (5/2 − ½i)(1 + i) = 5/2 + (5/2)i − ½i − ½i² = 5/2 + 2i + ½ = 3 + 2i. It works.

Worked example 2. Paper 1. Express (3 + 2i)/(1 − 4i) in the form a + bi.

(3 + 2i)/(1 − 4i) × (1 + 4i)/(1 + 4i)

= (3 + 12i + 2i + 8i²) / (1 + 16)

= (3 + 14i − 8) / 17

= −5/17 + (14/17)i

The denominator 1² + 4² = 17 comes straight from zz* = a² + b², so you can write it down without expanding. Leave exact fractions on Paper 1.

6Equating real and imaginary parts

Many Paper 1 questions give an equation in z and ask you to find z. Write z = a + bi (and z* = a − bi if it appears), simplify each side into the form (real) + (imaginary)i, and equate. You get two real equations in a and b.

Worked example 3. Paper 1. Find the complex number z such that 2z − 3z* = −1 + 10i.

let z = a + bi, so z* = a − bi

2(a + bi) − 3(a − bi) = −1 + 10i

2a + 2bi − 3a + 3bi = −1 + 10i

−a + 5bi = −1 + 10i

real parts: −a = −1, so a = 1

imaginary parts: 5b = 10, so b = 2

z = 1 + 2i

Square roots of a complex number. The same method finds the numbers whose square is a given complex number.

Worked example 4. Paper 1. Find the two complex numbers z such that z² = 5 + 12i.

let z = a + bi: (a + bi)² = a² − b² + 2abi = 5 + 12i

real parts: a² − b² = 5

imaginary parts: 2ab = 12, so b = 6/a

a² − 36/a² = 5

a⁴ − 5a² − 36 = 0

(a² − 9)(a² + 4) = 0

a² = 9, since a is real and a² = −4 is impossible

a = 3, b = 2 or a = −3, b = −2

z = 3 + 2i or z = −3 − 2i

The two answers are negatives of each other, as square roots always are: 5 + 12i is the z² you found in section 4. The key line is the one that rejects a² = −4 because a is real; say so, because that is the reasoning the marks are for.

7The complex plane

Because a + bi is fixed by two real numbers, it can be drawn as the point (a, b). The horizontal axis is the real axis and the vertical axis is the imaginary axis. This picture is the complex plane, also called the Argand diagram. Figure 1 plots some numbers.

Figure 1 · The complex plane (Argand diagram)

The real numbers are the horizontal axis, the familiar number line. The complex numbers fill the whole plane around it. That is the sense in which they generalize the reals: nothing was taken away, a dimension was added.

The operations you have learned have clean geometric meanings.

Conjugate and negative. z* = a − bi is the point (a, −b): the reflection of z in the real axis. −z = −a − bi is the point (−a, −b): a half-turn about the origin. Figure 2 shows both.

Figure 2 · z, its conjugate z* and its negative −z

Adding and subtracting. Think of z as the arrow from the origin to (a, b). Then complex numbers add exactly as vectors do (AHL 3.12): z + w is the fourth corner of the parallelogram on z and w. And z − w is the arrow from w to z, because w + (z − w) = z. Figure 3 draws both with z = 3 + 2i and w = 1 + i.

Figure 3 · Adding and subtracting on the complex plane

Multiplication also has a picture, and it is the reason the next subtopic exists. Multiplying by i turns every point a quarter-turn anticlockwise about the origin: i × (3 + 2i) = −2 + 3i, which is (3, 2) rotated by 90°. Figure 4 shows the powers of i doing exactly that. Subtopic 1.13 explains why, and what multiplying by any other complex number does.

Figure 4 · Powers of i go round in fours

8Modulus and argument

Two numbers describe where a point is: how far from the origin, and in what direction.

The modulus |z| = √(a² + b²) is the distance from the origin to z. The argument arg z is the angle from the positive real axis to the line joining the origin to z, measured anticlockwise.

Figure 5 shows both for z = 3 + 2i. The modulus comes from Pythagoras: |3 + 2i| = √(9 + 4) = √13. Note that |z|² = a² + b² = zz*, the product from section 5.

Figure 5 · Modulus and argument of z = 3 + 2i

The argument is measured in radians. It is usually given in the range −π < arg z ≤ π, the principal argument; a question may ask for 0 ≤ arg z < 2π instead, and will say so. Anticlockwise angles are positive, clockwise ones negative. Zero has no argument.

Finding the argument: draw the point first. tan(arg z) = b/a, but the calculator's arctan only returns angles between −π/2 and π/2, the first and fourth quadrants. For a point to the left of the imaginary axis it gives the wrong angle. So sketch where z is, find the acute angle the line makes with the real axis, and then place it in the right quadrant. Figure 6 does one in each quadrant.

Figure 6 · The argument in each quadrant

Worked example 5. Paper 1. Find the modulus and argument of z = −√3 + i.

|z| = √((−√3)² + 1²) = √(3 + 1) = 2

z is in the second quadrant (real part negative, imaginary part positive)

acute angle with the real axis: arctan(1/√3) = π/6

arg z = π − π/6 = 5π/6

The calculator would give arctan(1/(−√3)) = −π/6, which points into the fourth quadrant, the opposite direction. That is wrong by exactly π, and it is the most common way to lose the argument mark.

For −2 − 2i, in the third quadrant, the acute angle is π/4 and arg = −π + π/4 = −3π/4. For 3 − 4i, in the fourth quadrant, arctan(−4/3) is already correct: arg ≈ −0.927 radians (Paper 2, 3 s.f.), and the modulus is 5.

The modulus as a distance. Since z − w is the arrow from w to z, |z − w| is the distance between the points z and w. For z = 3 + 2i and w = 1 − 4i, |z − w| = |2 + 6i| = √40 = 2√10. In AHL 3 and in Paper 3, questions describe circles and other shapes this way: |z − 2i| = 3 is the circle of radius 3 centred at 2i.

9Technology

On Paper 2 the GDC can work in complex mode. It adds, multiplies and divides numbers typed as a + bi, and it has functions for the real part, the imaginary part, the conjugate, the modulus (usually abs) and the argument (usually angle or arg). Set it to radians. Use it to check Paper 2 arithmetic, but write down the expression you entered and the answer in the form asked for. On Paper 1 all of this must be done by hand, and exact answers such as 2√10 or 5π/6 are expected.

10Where marks are lost

Forgetting i² = −1. (3 + 2i)(1 + i) is 1 + 5i, not 3 + 5i + 2i². Every i² must become −1 before the answer is written.

Giving Im(z) as bi. For z = 5 − 2i, Im(z) = −2, a real number. Im(z) = −2i is wrong.

Dividing without the conjugate. (3 + 2i)/(1 + i) is not 3 + 2i. Multiply top and bottom by 1 − i. Multiplying by 1 + i instead leaves i in the denominator.

Taking the conjugate of the wrong thing. In division, use the conjugate of the denominator. In an equation containing z*, write z* = a − bi before you simplify.

The argument in the wrong quadrant. arctan(b/a) is only right when a > 0. Sketch the point, then adjust by π. For −√3 + i the answer is 5π/6, not −π/6.

Keeping impossible roots. When solving for a and b, a and b are real, so a² = −4 is rejected, and you must say why.

Decimals on Paper 1. |z| = √13 and arg z = 5π/6 are the answers. 3.61 or 2.62 lose the accuracy mark when an exact value was expected.

11Work it right

Put every complex number in a + bi form before you start, and simplify every i² to −1 immediately.
For division, multiply top and bottom by the conjugate of the denominator. The new denominator is a² + b²; write it down directly.
For an equation in z, write z = a + bi (and z* = a − bi), collect real and imaginary parts on each side, and equate them as two separate equations.
When solving for a and b, remember they are real, and reject any value that is not.
For an argument, sketch the point first, find the acute angle, then place it in the right quadrant. State the range you are using if the question does not.
Check a division by multiplying back, and a root by substituting it into the equation.
On Paper 1, give surds and multiples of π; on Paper 2, 3 significant figures in radians unless told otherwise.

12Try it

Marks in brackets. Q1 to Q4 and Q6 are Paper 1 style (no calculator). Q5 is Paper 2 style.

Q1. Let z = 2 − 3i and w = 4 + i.

(a) Find zw in the form a + bi. 2 marks

(b) Find z/w in the form a + bi, where a, b ∈ ℚ. 3 marks

Q2. Solve the equation z² − 6z + 34 = 0, giving your answers in the form a + bi. 3 marks

Q3. Find the complex number z such that 3z + 2iz* = 7 + 8i. 5 marks

Q4. Find the two complex numbers z that satisfy z² = −24 − 10i. 6 marks

Q5. Paper 2. Let z = −5 + 2i.

(a) Plot z and z* on an Argand diagram. 2 marks

(b) Find |z| and arg z, where −π < arg z ≤ π. 3 marks

Q6.

(a) Show that (1 + i)² = 2i. 1 mark

(b) Hence find the value of (1 + i)⁸. 2 marks

13In one breath

i² = −1, so √(−16) = 4i and every quadratic has roots, coming in pairs a ± bi when the discriminant is negative. A complex number z = a + bi has real part a and imaginary part b (a real number), and two complex numbers are equal only when both parts match, which turns one complex equation into two real ones. Add and multiply as in algebra, replacing i² by −1; divide by multiplying top and bottom by the conjugate of the denominator, since zz* = a² + b² is real. Plot a + bi at (a, b) on the complex plane: z* is the reflection in the real axis, −z the half-turn, z + w the parallelogram corner, and |z − w| the distance between z and w. The modulus |z| = √(a² + b²) is the distance from the origin, and the argument is the angle from the positive real axis, principal value in −π < arg z ≤ π; always sketch the point before taking arctan, because the calculator only knows two quadrants.

Answers

Q1. (a) zw = (2 − 3i)(4 + i) = 8 + 2i − 12i − 3i² = 8 − 10i + 3 = 11 − 10i. (b) z/w = (2 − 3i)(4 − i)/((4 + i)(4 − i)) = (8 − 2i − 12i + 3i²)/17 = (5 − 14i)/17 = 5/17 − (14/17)i. (c) |z| = √(4 + 9) = √13. (a) M1 for expanding with i² = −1, A1. (b) M1 for multiplying top and bottom by 4 − i, A1 for denominator 17, A1 for 5/17 − (14/17)i. (c) A1. Leaving 3i² unsimplified in (a) loses the A1.

Q2. z = [6 ± √(36 − 136)]/2 = [6 ± √(−100)]/2 = (6 ± 10i)/2, so z = 3 + 5i or z = 3 − 5i. M1 for the quadratic formula or completing the square ((z − 3)² = −25), A1 for √(−100) = 10i, A1 for both roots.

Q3. Let z = a + bi, z* = a − bi. Then 3a + 3bi + 2i(a − bi) = 3a + 3bi + 2ai − 2bi² = (3a + 2b) + (2a + 3b)i. Equate: 3a + 2b = 7 and 2a + 3b = 8. Multiply the first by 3 and the second by 2 and subtract: 5a = 5, so a = 1 and b = 2. z = 1 + 2i. M1 for substituting a + bi and a − bi, A1 for (3a + 2b) + (2a + 3b)i, M1 for equating real and imaginary parts, A1 for the two equations, A1 for z = 1 + 2i. Check: 3(1 + 2i) + 2i(1 − 2i) = 3 + 6i + 2i + 4 = 7 + 8i.

Q4. Let z = a + bi. Then a² − b² + 2abi = −24 − 10i, so a² − b² = −24 and ab = −5. Substitute b = −5/a: a² − 25/a² = −24, so a⁴ + 24a² − 25 = 0, (a² + 25)(a² − 1) = 0. a is real, so a² = 1: a = 1, b = −5 or a = −1, b = 5. z = 1 − 5i or z = −1 + 5i. M1 for expanding (a + bi)², M1 for equating both parts, A1 for the two equations, M1 for eliminating b to a quartic in a (or quadratic in a²), R1 for rejecting a² = −25 because a is real, A1 for both values of z. Check: (1 − 5i)² = 1 − 10i − 25 = −24 − 10i.

Q5. (a) z at (−5, 2), in the second quadrant; z* at (−5, −2), its reflection in the real axis. (b) |z| = √(25 + 4) = √29 ≈ 5.39. z is in the second quadrant, so arg z = π − arctan(2/5) ≈ 2.76 (radians). (a) A1 for each point correctly placed and labelled. (b) A1 for |z|, M1 for using the second quadrant (π − arctan(2/5), or a GDC's angle function), A1 for 2.76. An answer of −0.381 (the calculator's arctan(2/(−5))) scores M0 A0.

Q6. (a) (1 + i)² = 1 + 2i + i² = 1 + 2i − 1 = 2i, as required. (b) (1 + i)⁸ = ((1 + i)²)⁴ = (2i)⁴ = 16i⁴ = 16. (c) 2027 = 4 × 506 + 3, so i²⁰²⁷ = i³ = −i. (a) A1 with i² = −1 shown (AG). (b) M1 for (2i)⁴, A1 for 16. (c) A1. An answer of 16i in (b) comes from i⁴ taken as i, and scores M1 A0.

Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section AHL 1.12 Complex numbers: Cartesian form, the complex plane. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

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