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Year 12 · Complex Numbers · 22 min

Powers and roots

Find every nth root by accounting for all possible full turns.

01 / LEARN

The idea, step by step

One equation, several roots

For zn=R⁢cis⁢θ with R>0 and positive integer n, the modulus of each root is R1⁢/⁢n. Its angle must satisfy n⁢ϕ=θ+2⁢k⁢π. Thus the roots are R1⁢/⁢ncis⁢((θ+2⁢k⁢π)⁢/⁢n) for k=0,…,n−1.

Exactly n distinct roots arise: successive arguments differ by 2⁢π⁢/⁢n, making a regular polygon on a circle centred at the origin. Increasing k by n merely adds a full turn to a root. If the right side is zero, the sole distinct root is zero.

zk=R1⁢/⁢ncis(θ+2⁢k⁢πn),k=0,…,n−1

Do not divide only the principal angle

Dividing a single principal argument gives only one of the roots. The extra whole turns must be introduced before division. Afterwards, angles may be reduced to a requested interval.

A root is checked by raising its modulus to n and multiplying its angle by n. Listing the expected count, with distinct angles modulo 2⁢π, helps prevent omission or duplication.

A common trap

Writing only R1⁢/⁢ncis⁢(θ⁢/⁢n) as all the roots.

Instead: Add 2⁢k⁢π before dividing the angle by n, then select n distinct values.

02 / WORKED EXAMPLES

See the thinking

Example 1

Cube roots of a negative real number

Solve z3=−8.

  1. Write −8=8⁢cis⁢π.
  2. Each modulus is 2 and the arguments are (π+2⁢k⁢π)⁢/3 for k=0,1,2.
  3. These three angles are distinct modulo a full turn.
z=2⁢cis⁢(π⁢/3), 2⁢cis⁢π, 2⁢cis⁢(5⁢π⁢/3), or 1+3i,−2,1−3i.
Example 2

Four roots of a non-real number

Solve z4=16⁢i in polar form.

  1. 16⁢i=16⁢cis⁢(π⁢/2).
  2. The fourth-root modulus is 2; divide π⁢/2+2⁢k⁢π by 4.
  3. The angle step is π⁢/2, producing four distinct roots.
z=2⁢cis⁢(π⁢/8+k⁢π⁢/2), k=0,1,2,3.

Check your understanding

How many distinct roots does z5=32 have, and what is their common modulus?

Show answer

Five roots, each of modulus 2.

The nonzero right side gives five distinct equally spaced arguments; 321⁢/5=2.

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Read the questions online
  1. Solve z2=−9.
    2 marks
  2. List the cube roots of 8 in polar form.
    2 marks
  3. Find all solutions of z4=−16.
    3 marks
  4. Write the roots of z3=8⁢i in principal-argument polar form.
    3 marks
  5. Explain why k=4 gives no new root in zk=2⁢cis⁢(π⁢/8+k⁢π⁢/2).
    2 marks
  6. The three cube roots of a nonzero number form an equilateral triangle. Explain using arguments.
    3 marks
Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

  1. Solve z2=−9.
    2 marks

    Use arguments (π+2⁢k⁢π)⁢/2 for k=0,1 and modulus 3.

    z=3⁢i or z=−3⁢i.
  2. List the cube roots of 8 in polar form.
    2 marks

    The angles are 2⁢k⁢π⁢/3 for k=0,1,2.

    2, 2⁢cis⁢(2⁢π⁢/3), 2⁢cis⁢(4⁢π⁢/3).
  3. Find all solutions of z4=−16.
    3 marks

    The modulus is 2. Arguments come from (π+2⁢k⁢π)⁢/4.

    2⁢cis⁢(π⁢/4+k⁢π⁢/2), k=0,1,2,3.
  4. Write the roots of z3=8⁢i in principal-argument polar form.
    3 marks

    Angles from π⁢/6+2⁢k⁢π⁢/3 are π⁢/6,5⁢π⁢/6,3⁢π⁢/2; reduce the last by 2⁢π.

    2⁢cis⁢(π⁢/6), 2⁢cis⁢(5⁢π⁢/6), 2⁢cis⁢(−π⁢/2).
  5. Explain why k=4 gives no new root in zk=2⁢cis⁢(π⁢/8+k⁢π⁢/2).
    2 marks

    The angle for k=4 is π⁢/8+2⁢π, the same direction as π⁢/8.

    It repeats the k=0 root.
  6. The three cube roots of a nonzero number form an equilateral triangle. Explain using arguments.
    3 marks

    Equal central angles on one circle give equal chord lengths, so all three sides are equal.

    Adjacent roots differ in angle by 2⁢π⁢/3 and have equal modulus.
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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics complex numbers: Cartesian and polar representations, de Moivre, roots and roots of unity, Argand geometry and loci, complex factorisation and conjugate roots of real polynomials. Real-polynomial techniques are taught in the separate Real Polynomials chapter.

Official curriculum source ↗
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