Roots of unity — cube roots, nth roots, geometric interpretation: Question bank

Figure — Roots of unity — cube roots, nth roots, geometric interpretation

True or false — justify

True or false: every solution of

z^{n} = 1

has modulus

1

True — matching moduli in

r^{n} e^{in θ} = 1

forces

r^{n} = 1

, and since

r \geq 0

is real the only solution is

r = 1

, so all roots sit on the unit circle (the dashed circle in the figure above).

True or false: over the real numbers,

x^{4} = 1

has four solutions.

False — over

R

it has only two (

x = \pm 1

); the other two roots

\pm i

exist only because complex numbers carry direction (angle), which reals cannot.

True or false:

- 1

is a root of unity for every

n

False —

- 1 = e^{iπ}

is an

n

th root of unity only when

(- 1)^{n} = 1

, i.e. when

n

is even; for odd

n

(like the cube roots)

- 1

is not among them.

True or false: the sum of all

n

th roots of unity is

0

for every

n \geq 1

False — it is

0

only for

n \geq 2

. For

n = 1

the single root is

1

itself, and one vector cannot cancel, so the sum is

1

True or false: multiplying a complex number by

ω = e^{i 2 π / n}

changes its length.

False —

∣ ω ∣ = 1

, so multiplying by it is a pure rotation by

2 π / n

; lengths are preserved (see De Moivre's Theorem and the recap above).

True or false: the roots of unity are always symmetric about the real axis.

True — non-real roots come in conjugate pairs

ω^{k}

and

ω^{n - k} = \overline{ω^{k}}

, so the picture is a mirror image across the horizontal axis; the figure below shows this pairing explicitly.

True or false:

ω^{2}

(a cube root) equals

\overline{ω}

True — for the cube roots

ω^{2} = e^{i 4 π /3} = e^{- i 2 π /3} = \overline{ω}

; the "third" vertex is just the reflection of

ω

below the real axis (the top/bottom pair in the mirror figure).

True or false: every

n

th root of unity is a primitive

n

th root of unity.

False —

ω^{k}

is primitive only when

k

and

n

share no common factor bigger than

1

. For

n = 4

ω^{2} = - 1

shares the factor

2

with

4

, so its powers only ever hit

- 1

and

1

— never all four roots.

True or false: the product of all

n

th roots of unity is always

1

False — by Vieta the product is

(- 1)^{n + 1}

: it is

+ 1

for odd

n

and

- 1

for even

n

(e.g.

1 \cdot i \cdot (- 1) \cdot (- i) = - 1

Spot the error

Spot the error: "

z^{3} = 1

, therefore

z = 1

The cube function multiplies the angle by

3

; angles

0, \frac{2 π}{3}, \frac{4 π}{3}

all triple to a full turn, so there are three roots, not one. The figure below shows the three arrows folding onto

1

when cubed.

Spot the error: "The roots of

z^{n} = 1

are

θ = \frac{2 π k}{n}

for all integers

k

, so there are infinitely many."

Beyond

k = 0, \dots, n - 1

the angles differ by a multiple of

2 π

and land on the same points; distinct roots number exactly

n

Spot the error: "For cube roots

ω^{3} = ω

, since you 'come back around.'"

You come back to

1

, not to

ω

ω^{3} = 1

. Exponents reduce mod

n

, so

ω^{3} = ω^{0} = 1

, not

ω^{1}

Spot the error: "

\sum ω^{k} = \frac{ω ^{n} - 1}{ω - 1}

, and since

ω^{n} = 1

the top is

0

, but the bottom is also

0

, so the sum is undefined."

The bottom is

ω - 1

where

ω \neq = 1

(it is a different root), so

ω - 1 \neq = 0

; the fraction is a genuine

\frac{0}{nonzero} = 0

. The picture below shows why it must be zero: laid tip-to-tail the root vectors close into a loop back to the start. See Geometric Series.

Spot the error: "To solve

z^{3} = 8

, take

r^{3} = 8

r = 2

, and that's the only root:

z = 2

Modulus gives

r = 2

, but the argument still spreads over

\frac{2 π k}{3}

; the three roots are

2, 2 ω, 2 ω^{2}

— a triangle of radius

2

, not a single point.

Spot the error: "

1 + ω + ω^{2} + ω^{3} = 0

for cube roots, because roots sum to zero."

There are only three cube roots; the term

ω^{3} = 1

is not a fourth root but a repeat of

z_{0}

. The correct identity is

1 + ω + ω^{2} = 0

, so this sum equals

0 + 1 = 1

Spot the error: "The

n

th roots of unity form a regular

n

-gon centred at

1

The polygon is centred at the origin

0

; the vertex at

1

is one corner. Its centroid being

0

is exactly why the roots sum to zero (see Regular Polygons and the tip-to-tail loop above).

Why questions

Why must

z^{n} = 1

have exactly

n

solutions, no more, no fewer?

z^{n} - 1

is a degree-

n

polynomial, and by the Fundamental Theorem of Algebra a degree-

n

polynomial has

n

roots counted with multiplicity — and here all

n

are distinct (no repeated factors), so exactly

n

Why do we write

z

in polar form to solve

z^{n} = 1

In polar form a power just does

r^{n} e^{in θ}

— raising to a power scales the modulus and multiplies the angle (see the recap), turning a hard algebra problem into two easy conditions: match modulus, match angle.

Why is the sum of the roots geometrically "obviously" zero?

The

n

position vectors are spread at equal angles around the circle, so they form a balanced star; laid tip-to-tail they close into a loop (figure s04), and the centre of mass sits at the origin.

Why is

ω = e^{i 2 π / n}

called primitive?

Its successive powers

ω^{0}, ω^{1}, \dots, ω^{n - 1}

sweep through all

n

roots once each — one

\frac{2 π}{n}

step at a time generates the whole set, which a non-primitive root cannot do.

Why does the "no common factor" condition (

g cd (k, n) = 1

) decide primitivity?

Stepping by

ω^{k}

means jumping

k

vertices at a time; if

k

and

n

share a common factor

d > 1

you only ever land on every

d

-th vertex and cycle early, whereas coprime

k

visits all

n

vertices before repeating.

Why does

(1 + ω)

simplify to

- ω^{2}

for a cube root?

From

1 + ω + ω^{2} = 0

we isolate

1 + ω = - ω^{2}

; the identity lets us trade a sum for a single power, which is why cube-root problems collapse so cleanly.

Why does raising to the

n

th power "fold"

n

different angles onto the same result?

Multiplying an angle by

n

maps

\frac{2 π k}{n} \mapsto 2 π k

, and every full-turn multiple

2 π k

is the same direction (

1

); so

n

distinct starting angles all land on

1

after the

n

th power (this is exactly the folding shown in figure s03).

Why can't you find the roots of unity using only real square roots and arithmetic?

Real operations cannot produce rotation — you need the imaginary unit

i

to encode "turn by an angle," and rotation is exactly what raising to a power does on the Argand Diagram.

Edge cases

Edge case: what are the

n

th roots of unity when

n = 1

Just the single root

z = 1

— a "polygon" of one vertex; the sum is

1

(not

0

), which is why the sum-is-zero rule requires

n \geq 2

Edge case: what are the roots when

n = 2

1

and

- 1

— the degenerate "2-gon" is a straight line segment across the circle; they still sum to

0

, matching the general rule.

Edge case: does

z = 0

ever count as a root of unity?

Never —

0^{n} = 0 \neq = 1

for all

n

, and roots of unity must have modulus

1

, whereas

0

has modulus

0

; the origin is the centre, not a vertex.

Edge case: for a primitive root, which powers give back

1

Exactly the multiples of

n

ω^{k} = 1

iff

n ∣ k

. This is the smallest such

n

, which is precisely what "order

n

" and "primitive" mean.

Edge case: is

1

itself a primitive

n

th root of unity for any

n > 1

No —

1 = ω^{0}

has order

1

, so its powers are all

1

and never generate the other roots; the root

z_{0} = 1

is primitive only in the trivial case

n = 1

Edge case: what happens to the polygon as

n \to \infty

The vertices crowd ever closer together with spacing

\frac{2 π}{n} \to 0

, so the regular

n

-gon fills out and approaches the full unit circle — the figure below stacks

n = 4, 8, 16

so you can watch it happen.

Recall One-line summary to carry away

Roots of unity are $n$ equal dots on the unit circle, one at $1$ , centroid at the origin — every trap above is either forgetting there are $n$ of them, forgetting they live at radius $1$ , or forgetting exponents reduce mod $n$ .