Roots and identities
Find the th roots of a complex number; identify roots of unity and their properties; use De Moivre's theorem to derive trigonometric identities for and .
Worked examples
Finding cube roots of a complex number
Straightforward
Problem
Find the three cube roots of .
1
Write in exponential form.
, since and
2
Apply the th root formula with .
3
Compute each of the three roots.
: . : . :
4
Check one of the roots.
✓
Answer
The three cube roots of are , , and .
Deriving and using De Moivre's theorem
Moderate
Problem
Use De Moivre's theorem to express and in terms of and .
1
Apply De Moivre's theorem with .
2
Expand the left side.
3
Equate real and imaginary parts.
and
Answer
and .
Proving the sum of the roots of unity is zero
Challenging
Problem
Prove that , where .
1
Recognise the left side as a geometric series and apply the geometric sum formula.
The left side has first term , common ratio , and terms, so
2
Simplify using .
Since : , valid since for .
Answer
.
Practise
Q1·Straightforward
How many solutions does have in ?
Explanation
The equation is a degree-5 polynomial equation. By the fundamental theorem of algebra, it has exactly complex roots.
These are the 5th roots of unity: for .
On the Argand diagram they are equally spaced around the unit circle at angles .
These are the 5th roots of unity: for .
On the Argand diagram they are equally spaced around the unit circle at angles .
Q2·Straightforward
Find .
Explanation
All points on the unit circle have modulus 1. Every root of unity lies on the unit circle, so .
Q3·Straightforward
The 4th roots of unity are . Find the sum .
Explanation
This is a general result: the sum of all th roots of unity is always (for ). Geometrically, the roots are equally spaced around the unit circle, so their position vectors cancel.
Q4·Moderate
Find the argument of in degrees.
Explanation
is the principal cube root of unity (other than 1), located at on the unit circle.
Q5·Moderate
The equation has two solutions. Find the argument (in degrees) of the solution with the smallest positive argument.
Explanation
Write (since and ).
The square roots of are:
- : , argument
- : , argument
The smallest positive argument is .
Check: ✓
The square roots of are:
- : , argument
- : , argument
The smallest positive argument is .
Check: ✓
Q6·Moderate
The cube roots of all have the same modulus . Find .
Explanation
Taking moduli of both sides of :
All three cube roots of lie on a circle of radius centred at the origin, equally spaced at intervals.
All three cube roots of lie on a circle of radius centred at the origin, equally spaced at intervals.
Q7·Moderate
Find the sum of all cube roots of .
Explanation
The cube roots of are the solutions of .
By Vieta's formulas for , the sum of the three roots is:
You can verify directly. The three roots are , and their sum is:
By Vieta's formulas for , the sum of the three roots is:
You can verify directly. The three roots are , and their sum is:
Q8·Moderate
De Moivre's theorem states . Which expression gives ?
Explanation
Expanding:
Equating real parts with :
Equating imaginary parts with :
Note: option (c) is the expression for , not .
Equating real parts with :
Equating imaginary parts with :
Note: option (c) is the expression for , not .
Q9·Challenging
Use De Moivre's theorem to expand and hence express as a polynomial in .
Find the coefficient of in this polynomial (include the sign).
Find the coefficient of in this polynomial (include the sign).
Explanation
By the binomial theorem:
Real part (by De Moivre):
Substitute :
The polynomial is . The coefficient of is .
Real part (by De Moivre):
Substitute :
The polynomial is . The coefficient of is .
Q10·Challenging
The cube roots of include and two non-real roots. Find the imaginary part of the cube root with argument between and . Give your answer correct to 2 decimal places.
Explanation
Write .
The three cube roots of are:
- : argument , so
- : argument , so
- : argument , so
The root with argument between and (excluding the real root ) is .
Imaginary part: .
The three cube roots of are:
- : argument , so
- : argument , so
- : argument , so
The root with argument between and (excluding the real root ) is .
Imaginary part: .
Q11·Challenging
The 4th roots of have the form for . Find the real part of the root with argument . Give your answer correct to 2 decimal places.
Explanation
Write .
The four 4th roots of are:
For :
Real part .
The four roots are , , , — equally spaced at intervals on a circle of radius .
The four 4th roots of are:
For :
Real part .
The four roots are , , , — equally spaced at intervals on a circle of radius .
Q12·Challenging
Use De Moivre's theorem to prove that:
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
By De Moivre's theorem:
Expand the left side:
Equating imaginary parts with :
Substitute :
Expand the left side:
Equating imaginary parts with :
Substitute :
Q13·Challenging
Let be a primitive th root of unity, where is a positive integer. Prove that the sum of all th roots of unity is zero:
✎ Work this one through on paper — proofs are self-assessed.
Worked proof
The th roots of unity are where .
**Proof using the geometric series formula:**
The sum is a geometric series with first term , common ratio , and terms:
(We may divide by since for .)
But .
Therefore:
**Alternative (algebraic):** The th roots of unity satisfy . By Vieta's formulas for , the sum of all roots is .
**Proof using the geometric series formula:**
The sum is a geometric series with first term , common ratio , and terms:
(We may divide by since for .)
But .
Therefore:
**Alternative (algebraic):** The th roots of unity satisfy . By Vieta's formulas for , the sum of all roots is .
Open Math
Roots and identities
Complex numbers · MEX-12-02
Name:
Date:
Q1Straightforward
How many solutions does have in ?
Q2Straightforward
Find .
Q3Straightforward
The 4th roots of unity are . Find the sum .
Q4Moderate
Find the argument of in degrees.
Q5Moderate
The equation has two solutions. Find the argument (in degrees) of the solution with the smallest positive argument.
Q6Moderate
The cube roots of all have the same modulus . Find .
Q7Moderate
Find the sum of all cube roots of .
Q8Moderate
De Moivre's theorem states . Which expression gives ?
- A.
- B.
- C.
- D.
Q9Challenging
Use De Moivre's theorem to expand and hence express as a polynomial in .
Find the coefficient of in this polynomial (include the sign).
Find the coefficient of in this polynomial (include the sign).
Q10Challenging
The cube roots of include and two non-real roots. Find the imaginary part of the cube root with argument between and . Give your answer correct to 2 decimal places.
Q11Challenging
The 4th roots of have the form for . Find the real part of the root with argument . Give your answer correct to 2 decimal places.
Q12Challenging
Use De Moivre's theorem to prove that:
Q13Challenging
Let be a primitive th root of unity, where is a positive integer. Prove that the sum of all th roots of unity is zero:
Worked solutions and answers at openmath.au/year-12/extension-2/complex-numbers/roots-and-identities