IB Maths AA HL Topic 1 — Number & Algebra Paper 1 & 2 HL only ~10 min read

Roots of Complex Numbers

In real numbers, “the square root of 16” is just 4. End of story. But complex numbers play by different rules: 16 actually has two square roots (4 and −4), and a complex number can have any number of nth roots. In fact, every nonzero complex number has exactly n distinct nth roots, and they’re all spaced evenly around a circle on the Argand plane. The cube roots form an equilateral triangle, the fourth roots form a square, the fifth roots form a regular pentagon, and so on. This note is where De Moivre’s theorem really pays off — finding the nth roots of any complex number turns into a single formula plus a counter k that takes n values.

📘 What you need to know

Every nonzero complex number has exactly n distinct nth roots. The square roots come in 2s; the cube roots in 3s; the nth roots in ns.
Square roots via the algebraic method: let w = a + bi, expand w² = z, equate real and imaginary parts to form simultaneous equations.
De Moivre’s formula for nth roots: if z = r(cos θ + i sin θ), then the nth roots are
z^1/n = r^1/n[cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)] for k = 0, 1, 2, …, n−1.
In exponential form, this is z^1/n = r^1/n e^{i(θ + 2kπ)/n}.
This formula is not in the formula booklet — you must memorise it.
Geometric structure: the n roots form a regular n-sided polygon centred at the origin, with vertices on a circle of radius r^1/n.
Adjacent roots are separated by an angle of 2π/n — the polygon’s “wedge angle”.
Once you have one root, the others come from rotating by 2π/n repeatedly.

Square roots — the algebraic method

Before bringing De Moivre into the picture, here’s an elementary way to find the two square roots of a complex number. It works using nothing more than expanding brackets and solving simultaneous equations.

🧭 Recipe — square roots by simultaneous equations

Write the unknown root as w = a + bi, where a, b ∈ ℝ.
Square it: w² = (a + bi)² = a² + 2abi + b²i² = (a² − b²) + 2abi.
Set w² = z and equate real parts: a² − b² = Re(z).
Equate imaginary parts: 2ab = Im(z).
Solve the two simultaneous equations for a and b. Usually you make b the subject of the imaginary equation, substitute into the real one, and solve a quartic in a.
The two solutions w and −w are the square roots.

This algebraic method only works conveniently for square roots. For higher roots — cube roots, fourth roots, and beyond — the simultaneous equations get unmanageable, and De Moivre’s theorem is by far the better tool. Use the algebraic method when you want a quick Cartesian-form answer and don’t want to convert through polar.

How many nth roots does a complex number have?

Here’s the key fact: a nonzero complex number has exactly n distinct nth roots. Not one, not two — n. That’s a sharp contrast to real numbers, where 16 has only two real square roots and just one real cube root.

Number of nth roots every nonzero complex number has exactly n distinct nth roots in ℂ

🤔 Why so many?

Because adding a full revolution of 2π to the argument doesn’t change the complex number itself, but it does change the angle when you divide by n. Specifically, dividing 2π by n gives a step of 2π/n — exactly the spacing between adjacent nth roots. After n such steps, you’ve gone all the way around and started repeating; that’s why there are exactly n.

De Moivre’s formula for nth roots

The general formula for the nth roots of z = r(cos θ + i sin θ) follows from applying De Moivre with power 1/n, but with the extra trick of adding 2kπ to the argument first (since adding any multiple of 2π doesn’t change z). The result:

The nth roots of a complex number z^1/n = r^1/n[cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)]
for k = 0, 1, 2, …, n − 1 ✗ NOT in the formula booklet — must be memorised

In exponential form, the same formula is more compact:

In exponential form z^1/n = r^1/n · e^{i(θ + 2kπ)/n}, k = 0, 1, 2, …, n − 1

🧭 Recipe — finding all n nth roots

Convert z to polar form if it’s given in Cartesian: find r and θ.
Compute the modulus of each root: it’s r^1/n — the same for all n roots.
Compute the arguments by substituting k = 0, 1, 2, …, n−1 into (θ + 2kπ)/n.
Adjust each argument to lie in the required range (usually −π < θ ≤ π or 0 ≤ θ < 2π) by adding/subtracting 2π if needed.
Write each root in the requested form (polar, exponential, or Cartesian).

Quick check: the n arguments are spaced exactly 2π/n apart. So once you’ve computed the first one, you can find all the others by adding 2π/n repeatedly. Use this as a sanity check on your arithmetic.

The geometric picture — a regular polygon

Plot all n roots on an Argand diagram and you’ll see something elegant: they sit at the vertices of a regular n-sided polygon centred at the origin. Every vertex is on the same circle (radius r^1/n) and adjacent vertices are 2π/n apart.

The three cube roots of 8 form an equilateral triangle

This pattern generalises beautifully:

The nth roots of any complex number form: n = 2 → two diametrically opposite points (180° apart). n = 3 → equilateral triangle. n = 4 → square. n = 5 → regular pentagon. n = 6 → regular hexagon. And so on.

Roots of unity — a special case

The nth roots of 1 are called the nth roots of unity. They’re worth special attention because they have a few elegant properties — and they show up everywhere from signal processing to crystallography.

Since 1 = 1 · cis(0), the formula gives:

nth roots of unity cis(2kπ/n), k = 0, 1, …, n − 1

All n roots sit on the unit circle (modulus 1), with arguments equally spaced by 2π/n. Two surprising properties:

Properties of roots of unity: (1) the n roots sum to zero (because they balance perfectly around the origin). (2) 1 is always a root (corresponding to k = 0) — the other n−1 are non-real.

Worked examples

WE 1

Find square roots using simultaneous equations

Find both square roots of z = −5 + 12i, giving your answers in Cartesian form.

Step 1: Let w = a + bi be a square root w² = (a + bi)² = a² + 2abi + b²i² = (a² − b²) + 2abi Step 2: Set w² = z and equate parts real: a² − b² = −5 …① imag: 2ab = 12 → ab = 6 …② Step 3: From ②, b = 6/a; substitute into ① a² − (6/a)² = −5 a² − 36/a² = −5 a⁴ + 5a² − 36 = 0 (multiply through by a²) Step 4: Solve the quadratic in a² (a² + 9)(a² − 4) = 0 a² = −9 (reject, since a ∈ ℝ) or a² = 4 a = ±2 → b = 6/a = ±3 Step 5: Match signs (since ab = 6 > 0, a and b have same sign) w = 2 + 3i or w = −2 − 3i always check: (2+3i)² = 4 + 12i + 9i² = 4 + 12i − 9 = −5 + 12i ✓

WE 2

Find square roots using De Moivre

Find both square roots of z = 9i in polar form, with each argument in the range −π < θ ≤ π.

Step 1: Convert to polar form z = 9i sits on the positive Im axis |z| = 9, arg(z) = π/2 z = 9 cis(π/2) Step 2: Apply the nth-root formula with n = 2 z^(1/2) = 9^(1/2) cis((π/2 + 2kπ)/2) = 3 cis(π/4 + kπ), k = 0, 1 Step 3: Substitute k = 0 and k = 1 k = 0: 3 cis(π/4) k = 1: 3 cis(π/4 + π) = 3 cis(5π/4) 5π/4 is out of range, so subtract 2π: 5π/4 − 2π = −3π/4 3 cis(π/4) and 3 cis(−3π/4) notice the two roots are diametrically opposite — separated by π exactly, as expected for square roots

WE 3

Cube roots of a positive real number

Find all three cube roots of 27 in polar form, with arguments in the range −π < θ ≤ π.

Step 1: Convert to polar form 27 = 27 + 0i sits on positive real axis |z| = 27, arg(z) = 0 Step 2: Apply the formula with n = 3 z^(1/3) = 27^(1/3) cis((0 + 2kπ)/3) = 3 cis(2kπ/3), k = 0, 1, 2 Step 3: Substitute each value of k k = 0: 3 cis(0) = 3 k = 1: 3 cis(2π/3) k = 2: 3 cis(4π/3) → adjust: 4π/3 − 2π = −2π/3 3, 3 cis(2π/3), 3 cis(−2π/3) three cube roots — only ONE is real (the obvious 3); the other two are complex conjugates

WE 4

Cube roots of a complex number

Find all three cube roots of z = −8i, giving your answers in polar form with arguments in the range −π < θ ≤ π.

Step 1: Convert to polar form −8i sits on the negative Im axis |z| = 8, arg(z) = −π/2 Step 2: Apply the formula with n = 3 z^(1/3) = 8^(1/3) cis((−π/2 + 2kπ)/3) = 2 cis((−π/2 + 2kπ)/3), k = 0, 1, 2 Step 3: Substitute each k k = 0: 2 cis(−π/6) k = 1: 2 cis((−π/2 + 2π)/3) = 2 cis(3π/6/1) wait — let me redo k = 1: arg = (−π/2 + 2π)/3 = (3π/2)/3 = π/2 k = 2: arg = (−π/2 + 4π)/3 = (7π/2)/3 = 7π/6 7π/6 is out of range, so subtract 2π: 7π/6 − 2π = −5π/6 2 cis(−π/6), 2 cis(π/2), 2 cis(−5π/6) spacing check: −π/6 → π/2 is 2π/3 apart ✓, π/2 → −5π/6 (going the long way) is also 2π/3 ✓

WE 5

Fourth roots in Cartesian form

Find all four fourth roots of −16 in Cartesian form.

Step 1: Convert −16 to polar form |z| = 16, arg(z) = π Step 2: Apply the formula with n = 4 z^(1/4) = 16^(1/4) cis((π + 2kπ)/4) = 2 cis(π/4 + kπ/2), k = 0, 1, 2, 3 Step 3: Substitute each k k = 0: 2 cis(π/4) k = 1: 2 cis(3π/4) k = 2: 2 cis(5π/4) → adjust: 5π/4 − 2π = −3π/4 k = 3: 2 cis(7π/4) → adjust: 7π/4 − 2π = −π/4 Step 4: Convert each to Cartesian using exact trig values 2 cis(π/4) = √2 + √2 i 2 cis(3π/4) = −√2 + √2 i 2 cis(−3π/4) = −√2 − √2 i 2 cis(−π/4) = √2 − √2 i ±√2 ± √2 i (all four sign combinations) these four roots form a square, rotated 45° from the axes — each at distance 2 from origin

WE 6

Cube roots of unity and their sum

Find all three cube roots of unity (the cube roots of 1) in Cartesian form, and verify that their sum is zero.

Step 1: 1 in polar form |1| = 1, arg(1) = 0 Step 2: Apply the formula with n = 3 1^(1/3) = 1 · cis(2kπ/3), k = 0, 1, 2 Step 3: Compute each root k = 0: cis(0) = 1 k = 1: cis(2π/3) = −1/2 + (√3/2)i k = 2: cis(4π/3) = cis(−2π/3) = −1/2 − (√3/2)i Step 4: Sum the three roots 1 + (−1/2 + √3/2 i) + (−1/2 − √3/2 i) real: 1 − 1/2 − 1/2 = 0 imag: √3/2 − √3/2 = 0 three cube roots of 1: 1, −1/2 ± (√3/2)i — sum = 0 ✓ this is a famous classical result — the n nth roots of unity always sum to zero (true for any n ≥ 2)

💡 Top tips

Memorise the nth-root formula — it’s not in the booklet. Practise writing r^1/n cis((θ + 2kπ)/n) until it’s automatic.
Always use values k = 0, 1, 2, …, n−1. Beyond k = n−1, the roots start repeating — don’t list extras.
The n roots are evenly spaced by 2π/n. Use this as a sanity check on your arguments — adjacent ones must differ by exactly 2π/n.
Plot the roots on an Argand diagram to confirm the regular polygon. Visual confirmation is fast and catches arithmetic errors.
For square roots, the algebraic method is often quicker than De Moivre — fewer steps and a Cartesian-form answer drops out directly.
For higher roots (cube and beyond), use De Moivre. The simultaneous-equations approach explodes in complexity.
Roots of unity are the cleanest case — they sit on the unit circle, sum to zero, and 1 is always one of them.

⚠ Common mistakes

Using only one root. A complex number has n distinct nth roots — list all of them, not just one.
Going beyond k = n−1. Values like k = n just produce repeats of earlier roots. Stop at k = n−1.
Forgetting to add 2kπ when applying the formula. Without that term, you only get one of the roots — the other n−1 are missing.
Mishandling out-of-range arguments. If a computed argument falls outside the question’s range, add or subtract 2π. Don’t just leave it.
Trying to use the algebraic (simultaneous-equations) method for cube or higher roots. It works in principle but the algebra becomes unmanageable. Use De Moivre instead.
Forgetting that the modulus of every root is the same. All n roots have modulus r^1/n — only the arguments differ.
Sign errors converting back to Cartesian. Watch the sign of cos and sin in each quadrant — Q3 has both negative, Q4 has positive cos and negative sin.

You’ve now reached the end of the Further Complex Numbers section — and effectively the end of the complex-number content in IB Maths AA HL. The journey has taken you from “x² = −1 has no solutions” all the way to nth roots and roots of unity sitting elegantly on regular polygons. Every result here builds on the same core idea: complex numbers extend real numbers by adding rotation. Once you can rotate, you can multiply (scale + rotate), take powers (rotate n times), and find roots (split a rotation into n equal parts). It’s one of the most beautiful arcs in the whole HL syllabus — congratulations on reaching the end of it.

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