Write the conjugate of z = -2 + 5i and evaluate zz. [2 marks]

SQA AH style: polar form and power-style practice: Write z = 1 + i in polar form and use de Moivre's theorem to find z^8.

Modulus |z| = sqrt(1^2 + 1^2) = sqrt(2); argument z = tan^-111 = π4 (2 marks). Polar form: z = sqrt(2)(cosπ4 + isinπ4) (1 mark). De Moivre: z^8 = (sqrt(2))^8(cos8π4 + isin8π4) = 16(cos 2π + isin 2π) (1 mark). = 16(1 + 0i) = 16 (1 mark). Markers reward the modulus and argument, the polar form, applying de Moivre, and the simplified real answer.

SQA AH style: division-style practice: Express 3 + 2i1 - i in the form a + bi.

Multiply top and bottom by the conjugate of the denominator, 1 + i (1 mark). Numerator: (3 + 2i)(1 + i) = 3 + 3i + 2i + 2i^2 = 3 + 5i - 2 = 1 + 5i. Denominator: (1 - i)(1 + i) = 1 - i^2 = 2 (1 mark). So 1 + 5i2 = 12 + 52i (1 mark). Markers reward multiplying by the conjugate, expanding both products using i^2 = -1, and the final form.

ScotlandMathsSyllabus dot point

How do you work with complex numbers in Cartesian and polar form, and use de Moivre's theorem for powers and roots?

Perform arithmetic with complex numbers in Cartesian form, represent them on an Argand diagram, convert to polar (modulus-argument) form, and use de Moivre's theorem to find powers and the nth roots of a complex number.

A focused answer to the SQA Advanced Higher Mathematics complex numbers content, covering arithmetic in Cartesian form, the complex conjugate, the Argand diagram, modulus and argument, polar form, and de Moivre's theorem for finding powers and the nth roots of a complex number.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Arithmetic in Cartesian form
The Argand diagram and polar form
De Moivre's theorem
Solving equations and the conjugate root pair
Try this

What this dot point is asking

The SQA wants you to do arithmetic with complex numbers, plot them on an Argand diagram, convert between Cartesian and polar form, and use de Moivre's theorem to raise complex numbers to powers and extract roots. This brings algebra and geometry together.

Arithmetic in Cartesian form

Treat $i$ as a symbol with $i^2 = -1$ . Addition and subtraction are componentwise; multiplication is ordinary expansion followed by replacing $i^2$ ; division uses the conjugate to make the denominator real.

Multiplying two complex numbers in Cartesian form

Multiply $(2 + 3i)(1 - 4i)$ .

Step 1: Expand the brackets

Multiply each term in the first bracket by each term in the second, exactly as with ordinary brackets. This produces four terms, one of which contains $i^2$ :

(2 + 3i)(1 - 4i) = 2 - 8i + 3i - 12i^2.

Step 2: Replace $i^2 = -1$

The defining property $i^2 = -1$ turns what looks like an imaginary term into a real one. This is the key step that keeps complex arithmetic in the form $a + bi$ :

= 2 - 5i - 12(-1) = 2 - 5i + 12.

Step 3: Collect real and imaginary parts

Group the real terms and the imaginary terms separately:

= 14 - 5i.

Final answer: $(2 + 3i)(1 - 4i) = 14 - 5i$ .

The Argand diagram and polar form

The Argand diagram plots $z = a + bi$ as the point $(a, b)$ . Its distance from the origin is the modulus and the angle from the positive real axis is the argument, which lets you rewrite $z$ in polar form.

The argument must be placed in the correct quadrant: $\tan^{-1}\dfrac{b}{a}$ alone does not distinguish, for example, the first and third quadrants, so always sketch the point first.

De Moivre's theorem

De Moivre's theorem turns powers and roots of complex numbers into simple arithmetic on the modulus and argument: raise the modulus to the power and multiply the argument by it.

Finding the three cube roots of

8

using de Moivre

Find the three cube roots of $8$ .

Step 1: Write $8$ in polar form, adding all equivalent arguments

$8$ is a positive real number with modulus $8$ and argument $0$ . But the argument is only defined up to multiples of $2\pi$ , and we need exactly three distinct values, so we write:

8 = 8(\cos 2k\pi + i\sin 2k\pi), \quad k = 0, 1, 2.

Step 2: Apply de Moivre's theorem with $n = \tfrac{1}{3}$

Take the cube root by raising to the power $\tfrac{1}{3}$ . De Moivre says to take the $\tfrac{1}{3}$ power of the modulus and divide the argument by $3$ :

8^{1/3}\left(\cos\frac{2k\pi}{3} + i\sin\frac{2k\pi}{3}\right) = 2\left(\cos\frac{2k\pi}{3} + i\sin\frac{2k\pi}{3}\right), \quad k = 0, 1, 2.

Step 3: Evaluate each of the three roots

Substitute $k = 0, 1, 2$ in turn:

$k = 0$ : $2(\cos 0 + i\sin 0) = 2$ .

$k = 1$ : $2\!\left(-\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}\right) = -1 + i\sqrt{3}$ .

$k = 2$ : $2\!\left(-\dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}\right) = -1 - i\sqrt{3}$ .

Final answer: the three cube roots of $8$ are $2$ , $-1 + i\sqrt{3}$ , and $-1 - i\sqrt{3}$ , equally spaced around a circle of radius $2$ .

Solving equations and the conjugate root pair

Complex numbers complete the solution of polynomial equations: every quadratic with a negative discriminant has a conjugate pair of complex roots. A key fact for real-coefficient polynomials is that complex roots always come in conjugate pairs, so if $z = a + bi$ is a root then $\bar{z} = a - bi$ is also a root. This lets you reconstruct a quadratic factor $(z - z_0)(z - \bar{z}_0)$ with real coefficients, which is the link back to the irreducible quadratics you met in partial fractions.

Solving a quadratic with complex roots

Solve $z^2 - 4z + 13 = 0$ .

Step 1: Apply the quadratic formula

The discriminant is $16 - 52 = -36 < 0$ , so the roots are complex. Apply the formula directly:

z = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{-36}}{2}.

Step 2: Simplify the square root of a negative

Write $\sqrt{-36} = \sqrt{36}\sqrt{-1} = 6i$ , then substitute:

z = \frac{4 \pm 6i}{2} = 2 \pm 3i.

Step 3: State the conjugate pair

For a polynomial with real coefficients, complex roots always appear in conjugate pairs. The two roots here, $2 + 3i$ and $2 - 3i$ , confirm this.

Final answer: the roots are $z = 2 + 3i$ and $z = 2 - 3i$ .

Try this

Q1. Find $|3 - 4i|$ . [1 mark]

Cue. $\sqrt{3^2 + (-4)^2} = \sqrt{25} = 5$ .

Q2. Write the conjugate of $z = -2 + 5i$ and evaluate $z\bar{z}$ . [2 marks]

Cue. $\bar{z} = -2 - 5i$ , and $z\bar{z} = (-2)^2 + 5^2 = 29$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AH style: polar form and power5 marksWrite

z = 1 + i

in polar form and use de Moivre's theorem to find

z^{8}

Show worked answer →

Modulus $|z| = \sqrt{1^2 + 1^2} = \sqrt{2}$ ; argument $\arg z = \tan^{-1}\dfrac{1}{1} = \dfrac{\pi}{4}$ (2 marks).

Polar form: $z = \sqrt{2}\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right)$ (1 mark).

De Moivre: $z^{8} = (\sqrt{2})^{8}\left(\cos\dfrac{8\pi}{4} + i\sin\dfrac{8\pi}{4}\right) = 16(\cos 2\pi + i\sin 2\pi)$ (1 mark).

$= 16(1 + 0i) = 16$ (1 mark). Markers reward the modulus and argument, the polar form, applying de Moivre, and the simplified real answer.

AH style: division3 marksExpress

\dfrac{3 + 2i}{1 - i}

in the form

a + bi

Show worked answer →

Multiply top and bottom by the conjugate of the denominator, $1 + i$ (1 mark).

Numerator: $(3 + 2i)(1 + i) = 3 + 3i + 2i + 2i^2 = 3 + 5i - 2 = 1 + 5i$ . Denominator: $(1 - i)(1 + i) = 1 - i^2 = 2$ (1 mark).

So $\dfrac{1 + 5i}{2} = \dfrac{1}{2} + \dfrac{5}{2}i$ (1 mark). Markers reward multiplying by the conjugate, expanding both products using $i^2 = -1$ , and the final form.

Related dot points

Sources & how we know this

SQA Advanced Higher Mathematics Course Specification — SQA (2019)