MathsISC Class 11 · Complex Numbers

Polar Form, Square Root of a Complex Number and Cube Roots of Unity

Q: What is the argument of 0?

It is undefined: the modulus is 0 and the direction is not determined.

Q: Why do the cube roots of unity form an equilateral triangle?

They lie on the unit circle at angles 0, 120° and 240°.

Q: How many square roots does a complex number have?

Two, one the negative of the other.

Updated 17 October 2026 · Fact-checked

In the Argand plane z = a + ib is the point (a, b). The polar form is z = r(cos θ + i sin θ) with r = |z| and θ the argument. The square root of a + ib is found by solving (x + iy)² = a + ib. The cube roots of unity are 1, ω, ω² with ω = (−1 + i√3)/2, 1 + ω + ω² = 0 and ω³ = 1.

Understand Polar Form, Square Roots and Cube Roots of Unity

Argand plane. The complex number $a + ib$ is the point $(a, b)$ ; the modulus is the distance from the origin and the argument θ is the angle from the positive real axis. The principal argument lies in $(- π, π]$ . To find θ, find the reference angle $α = tan^{- 1} ∣ b / a ∣$ and place it in the correct quadrant: first quadrant $θ = α$ ; second $θ = π - α$ ; third $θ = - (π - α)$ ; fourth $θ = - α$ .

Polar form. $z = r (cos θ + i sin θ)$ , with $r = a^{2} + b^{2}$ . Product: moduli multiply and arguments add. Quotient: moduli divide and arguments subtract. Powers (De Moivre's theorem): $[r (cos θ + i sin θ)]^{n} = r^{n} (cos n θ + i sin n θ)$ .

**Square root of $a + ib$ .** Let $(x + i y)^{2} = a + ib$ , so $x^{2} - y^{2} = a$ and $2 x y = b$ . Use also $x^{2} + y^{2} = a^{2} + b^{2}$ to get $x^{2}$ and $y^{2}$ . The sign of $x y$ follows the sign of b, and there are two roots, one the negative of the other.

Cube roots of unity. Solving $z^{3} = 1$ gives $1$ , $ω = \frac{- 1 + i 3}{2}$ and $ω^{2} = \frac{- 1 - i 3}{2}$ . Properties: $1 + ω + ω^{2} = 0$ ; $ω^{3} = 1$ ; $ω^{2} = \overset{ω}{ˉ} = 1/ ω$ ; $∣ ω ∣ = 1$ . Hence $ω^{n}$ depends on n modulo 3. Useful: $1 + ω = - ω^{2}$ , $1 + ω^{2} = - ω$ .

Factorisations. $x^{3} - 1 = (x - 1) (x - ω) (x - ω^{2})$ and $x^{2} + x + 1 = (x - ω) (x - ω^{2})$ . The three roots lie at the vertices of an equilateral triangle on the unit circle.

Polar Form, Square Roots and Cube Roots of Unity: formulas to remember

Polar form: $$z = r(\cos θ + i \sin θ)$ | $r = \sqrt{a^2 + b^2}$ | $θ = \tan^{-1}(b/a)$ (quadrant)$; Principal argument in $(- π, π]$ .
De Moivre: $[\cos θ + i \sin θ]^n = \cos nθ + i \sin nθ$; Integer n.
Cube roots of unity: $1$ | $ω = \frac{-1 + i\sqrt{3}}{2}$ | $ω^2 = \frac{-1 - i\sqrt{3}}{2}$; ω³ = 1.
Key identities: $$1 + ω + ω^2 = 0$ | $ω^3 = 1$ | $1 + ω = -ω^2$$; Reduce powers modulo 3.

Polar Form, Square Roots and Cube Roots of Unity in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Rotating a point by 90 degrees

Suppose a robot arm at the complex position $z = 3 + 4 i$ (metres) rotates about the origin through 90° anticlockwise.

Find the new position and show that the distance from the origin is unchanged.

Try it first, then show the solutionSolution

Rotation through θ multiplies z by $cos θ + i sin θ$ ; for 90° this is $i$ .
New position $= i (3 + 4 i) = 3 i + 4 i^{2} = - 4 + 3 i$ .
$∣ z ∣ = 9 + 16 = 5$ and $∣ - 4 + 3 i ∣ = 16 + 9 = 5$ .

Answer: −4 + 3i; the distance 5 m is unchanged.

Memory hook: Multiplying by i rotates by 90°.

Tricks to remember Polar Form, Square Roots and Cube Roots of Unity

Short memory aids. Say each one aloud once, then use it on a question the same day.

Quadrant first
Plot the point before choosing the argument.
Use it when: Finding θ.
Powers of ω mod 3
ω³ = 1, so ω^{3k+r} = ω^r.
Use it when: Evaluating high powers of ω.

Polar Form, Square Roots and Cube Roots of Unity: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Sum of cube roots · 1 mark · MCQBoard level

If ω is a complex cube root of unity, $1 + ω + ω^{2}$ equals:
(a) 1
(b) 0
(c) ω
(d) −1

Try it first, then show the solutionSolution

The three cube roots of unity sum to zero.

Answer: (b) 0

Example 2: Polar form · 3 marks · Short answerBoard level

Express $- 1 + i$ in polar form.

Try it first, then show the solutionSolution

$r = 1 + 1 = 2$ . (1 mark)
The point (−1, 1) lies in the second quadrant; reference angle $tan^{- 1} 1 = \frac{π}{4}$ , so $θ = π - \frac{π}{4} = \frac{3 π}{4}$ . (1 mark)
$- 1 + i = 2 (cos \frac{3 π}{4} + i sin \frac{3 π}{4})$ . (1 mark)

Answer: √2 (cos 3π/4 + i sin 3π/4).

Example 3: Square root · 4 marks · Short answerHarder board level

Find the square roots of $3 + 4 i$ and of $- 5 + 12 i$ .

Try it first, then show the solutionSolution

For $3 + 4 i$ : $∣ z ∣ = 5$ , $x^{2} = \frac{5 + 3}{2} = 4$ and $y^{2} = \frac{5 - 3}{2} = 1$ ; $x y > 0$ , so roots $\pm (2 + i)$ . (2 marks)
For $- 5 + 12 i$ : $∣ z ∣ = 13$ , $x^{2} = \frac{13 - 5}{2} = 4$ and $y^{2} = \frac{13 + 5}{2} = 9$ ; $x y > 0$ , so $\pm (2 + 3 i)$ . (1.5 marks)
Check: $(2 + i)^{2} = 3 + 4 i$ and $(2 + 3 i)^{2} = - 5 + 12 i$ . (0.5 mark)

Answer: ±(2 + i); ±(2 + 3i).

Example 4: De Moivre's theorem · 4 marks · Short answerHarder board level

Evaluate $(1 + i 3)^{6}$ using polar form, and find $ω^{25} + ω^{50} + 1$ .

Try it first, then show the solutionSolution

$1 + i 3$ has $r = 2$ and $θ = \frac{π}{3}$ . (1 mark)
$(2)^{6} (cos 2 π + i sin 2 π) = 64$ . (1.5 marks)
$ω^{25} = ω^{1}$ since $25 = 3 \times 8 + 1$ and $ω^{50} = ω^{2}$ since $50 = 3 \times 16 + 2$ . (1 mark)
So $ω + ω^{2} + 1 = 0$ . (0.5 mark)

Answer: 64; 0.

Example 5: Identities with ω · 5 marks · Long answerHOTS / competency

If ω is a non-real cube root of unity, show that $(1 - ω + ω^{2}) (1 + ω - ω^{2}) = 4$ and find the value of $(1 + ω)^{3} + (1 + ω^{2})^{3}$ .

Try it first, then show the solutionSolution

$1 + ω^{2} = - ω$ , so $1 - ω + ω^{2} = - 2 ω$ . (1 mark)
$1 + ω = - ω^{2}$ , so $1 + ω - ω^{2} = - 2 ω^{2}$ . (1 mark)
Product $= 4 ω^{3} = 4$ . (1 mark)
$(1 + ω)^{3} = (- ω^{2})^{3} = - ω^{6} = - 1$ and $(1 + ω^{2})^{3} = (- ω)^{3} = - ω^{3} = - 1$ . (1.5 marks)
Sum $= - 2$ . (0.5 mark)

Answer: 4; −2.

Where students lose marks in Polar Form, Square Roots and Cube Roots of Unity

The slips examiners see most often, and what to write instead.

Mistake: Wrong quadrant for the argument
Giving π/4 for −1 + i.
Do this: Locate the point, then adjust the reference angle.
Usually costs: 2 marks.
Mistake: Sign error in the square root
Taking xy > 0 when b < 0.
Do this: The sign of xy follows the sign of b.
Usually costs: 1 to 2 marks.
Mistake: Not reducing powers of ω
Computing ω^{100} directly.
Do this: Reduce the exponent modulo 3.
Usually costs: 1 mark.

Exam tips

Plot the point to decide the quadrant.
Check a square root by squaring.
Use 1 + ω + ω² = 0 to simplify.
Quote De Moivre's theorem for powers.

Practice questions on Polar Form, Square Roots and Cube Roots of Unity

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The modulus of $- 3 + 4 i$ and its principal argument quadrant are:
(a) 5, first
(b) 5, second
(c) 7, second
(d) 1, third
Hint
a < 0 and b > 0.
Answer
(b) 5, second
Q2 · 2 marks · Short answerBoard level
Write $1 - i$ in polar form.
Hint
Fourth quadrant.
Answer
$r = 2$ , $θ = - \frac{π}{4}$ : $2 (cos \frac{π}{4} - i sin \frac{π}{4})$ .
Q3 · 3 marks · Short answerHarder board level
Find the square root of $- 15 - 8 i$ .
Hint
$∣ z ∣ = 17$ .
Answer
$x^{2} = \frac{17 - 15}{2} = 1$ , $y^{2} = \frac{17 + 15}{2} = 16$ ; $x y < 0$ : roots $\pm (1 - 4 i)$ . Check: $(1 - 4 i)^{2} = 1 - 8 i - 16 = - 15 - 8 i$ .
Q4 · 3 marks · Short answerHarder board level
Show that $ω^{2} = \overset{ω}{ˉ}$ and evaluate $ω^{100} + ω^{200}$ .
Hint
ω² = (−1 − i√3)/2.
Answer
$ω = \frac{- 1 + i 3}{2}$ and $ω^{2} = \frac{- 1 - i 3}{2} = \overset{ω}{ˉ}$ . $ω^{100} = ω$ (100 = 99 + 1) and $ω^{200} = ω^{2}$ (200 = 198 + 2), so the sum is $- 1$ .
Q5 · 3 marks · Short answerHarder board level
Express $(1 + i)^{8}$ in the form $a + ib$ and give its modulus.
Hint
|1 + i| = √2, arg π/4.
Answer
$(2)^{8} (cos 2 π + i sin 2 π) = 16$ . Modulus 16.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose three signals have phasors at the cube roots of unity.
(i) Write the three phasors. (1 mark)
(ii) Find their sum. (1 mark)
(iii) Find the modulus of each. (1 mark)
(iv) Find the angle between consecutive phasors. (1 mark)
Hint
Roots of z³ = 1.
Answer
(i) $1$ , $ω = - 1/2 + (\sqrt3/2) i$ , $ω^{2} = - 1/2 - (\sqrt3/2) i$ . (ii) 0. (iii) 1 each. (iv) 120° (2π/3).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.ω³?Show answer
1.
2.1 + ω + ω²?Show answer
0.
3.Modulus of ω?Show answer
1.
4.Argument of −1?Show answer
π.

Polar Form, Square Roots and Cube Roots of Unity: frequently asked questions

What is the argument of 0?

It is undefined: the modulus is 0 and the direction is not determined.

Why do the cube roots of unity form an equilateral triangle?

They lie on the unit circle at angles 0, 120° and 240°.

How many square roots does a complex number have?

Two, one the negative of the other.