De Moivre's theorem — statement, proof, applications: Question bank

A quick shared vocabulary so nothing below is a surprise:

True or false — justify

Every statement is either exactly right or subtly broken. Decide, then justify.

TRUE or FALSE:

(cos θ + i sin θ)^{n} = cos (n θ) + i sin (n θ)

holds for every real number

n

False. Guaranteed only for integer

n

. For fractional

n

the left side is multivalued (a

q

-th root has

q

values), so a single clean equation cannot capture it.

TRUE or FALSE:

(cos θ - i sin θ)^{n} = cos (n θ) - i sin (n θ)

True.

cos θ - i sin θ = cos (- θ) + i sin (- θ) = cis (- θ)

, so its

n

-th power is

cis (- n θ) = cos n θ - i sin n θ

TRUE or FALSE: For

r > 0

(r (cos θ + i sin θ))^{n} = r^{n} (cos n θ + i sin n θ)

True. The modulus and the angle behave independently: the length is raised to the power

r^{n}

, the angle is multiplied

n θ

TRUE or FALSE: De Moivre's proof for positive integers needs Euler's formula

e^{i θ} = cis θ

False. Induction only needs the angle-addition formulas. Euler gives a slick one-liner but is not required — it is the "cheat", not the "earn".

TRUE or FALSE: The

n

n

-th roots of any complex number always sum to

0

False in general — true only when there are at least $2$ roots and no shift. For

z^{n} = w

with

n \geq 2

the roots are symmetric about the origin and sum to

0

; but

n = 1

gives a single root equal to

w

, whose "sum" is

w \neq = 0

unless

w = 0

TRUE or FALSE:

(2 cis θ)^{3} = 6 cis (3 θ)

False. The modulus is raised to the power, not multiplied by

n

(2)^{3} = 8

, giving

8 cis (3 θ)

TRUE or FALSE: If

z^{5} = 1

, then all five solutions lie on a circle of radius

1

True. Each root has modulus

1^{1/5} = 1

, so all lie on the unit circle, equally spaced by

2 π /5

. See Roots of unity.

TRUE or FALSE: Because

cis (2 π) = 1

, the angle

2 π

and the angle

0

describe the same complex number, so we can never distinguish them.

True for the number, but they matter for roots. Adding

2 π

leaves

w

unchanged, yet after dividing by

n

the shift

2 π / n

produces a genuinely different root — which is exactly why there are

n

of them.

Spot the error

Each line contains one flawed step. Name it.

Student writes

(1 + i)^{8} = (2)^{8} cis (8 \cdot \frac{π}{4}) = 16 cis (2 π) = 16 i

. Where is the slip?

cis (2 π) = cos 2 π + i sin 2 π = 1

, not

i

. The answer is

16

; they confused the full turn with a quarter turn.

Student claims

(cis θ)^{1/2} = cis (θ /2)

is the square root of

cis θ

. What is missing?

The second root. Fractional powers are multivalued; the two square roots are

cis (θ /2)

and

cis (θ /2 + π)

. Writing "the" root drops half the answer.

To find the cube roots of

8 i

, a student uses

ρ = 8^{1/3} = 2

but forgets the

2 π k

term, writing only

z_{0} = 2 cis (π /6)

. Consequence?

They report one root instead of three. The angle must be

\frac{π /2 + 2 π k}{3}

for

k = 0, 1, 2

, giving three equally spaced roots.

Student expands

(cos θ + i sin θ)^{2} = cos^{2} θ + i^{2} sin^{2} θ = cos^{2} θ - sin^{2} θ

and stops. Error?

They dropped the cross term

2 i cos θ sin θ

. The full result is

(cos^{2} θ - sin^{2} θ) + i (2 cos θ sin θ)

, i.e.

cos 2 θ + i sin 2 θ

Student says "

cis α \cdot cis β = cis (α β)

because we multiply." Fix it.

When arrows are multiplied, the angles add, not multiply:

cis α \cdot cis β = cis (α + β)

. Angles multiply only when you take powers.

Student proves the negative-power case by writing

(cis θ)^{- m} = - cis (m θ)

. What went wrong?

A negative exponent means reciprocal, not negation. Since

∣ cis (m θ) ∣ = 1

, its reciprocal is its conjugate

cis (- m θ)

, not

- cis (m θ)

Why questions

Ask "why" until the mechanism is exposed.

Why does the induction step in the proof rely specifically on the angle-addition formulas?

Because the step multiplies

cis (k θ)

cis θ

, and the only fact that turns a product of two cis's into a single cis is that

cos

and

sin

of a sum split via angle-addition.

Why must we add

2 π k

(and not, say,

π k

) when extracting

n

-th roots?

Only a full turn of

2 π

returns an angle to the same complex number. Adding

π

would flip to a different point, so it would not be a legitimate relabelling of

w

Why are there exactly

n

distinct

n

-th roots — not fewer, not infinitely many?

Dividing the shifts

2 π k

n

gives angles

2 π k / n

that stay distinct for

k = 0, \dots, n - 1

; at

k = n

the angle has advanced by a full

2 π

and coincides with

k = 0

, so the list closes after

n

terms.

Why can De Moivre generate multiple-angle identities like

cos 3 θ = 4 cos^{3} θ - 3 cos θ

Expanding

(cos θ + i sin θ)^{n}

with the Binomial theorem gives one expression; De Moivre says it also equals

cos n θ + i sin n θ

. Matching real parts links a single multiple-angle to powers of

cos

and

sin

Why do we equate real and imaginary parts separately when deriving these identities?

Two complex numbers are equal iff their real parts match and their imaginary parts match — they live on independent axes of the Argand diagram, so one equation actually carries two.

Why does raising to a power stretch the arrow rather than leave its length alone?

A power is repeated multiplication, and multiplying moduli multiplies lengths; doing this

n

times turns length

r

into

r^{n}

Why is the reciprocal of

cis (m θ)

so easy to find?

Its modulus is

1

, so it sits on the unit circle; for unit-modulus numbers the inverse equals the conjugate, which just flips the angle's sign to

cis (- m θ)

Edge cases

The scenarios people forget to check.

What does De Moivre give when

n = 0

(cis θ)^{0} = 1 = cos 0 + i sin 0 = cis (0 \cdot θ)

. The theorem still holds — every arrow to the zeroth power is the unit arrow at angle

0

What happens to the

n

-th-root formula when

w = 0

All roots collapse to the single point

0

: modulus

0^{1/ n} = 0

, so there is no circle and no spacing — the "equally spaced" picture degenerates to the origin.

What if

θ = 0

(a positive real number) is raised to the power

n

cis 0 = 1

, and

1^{n} = 1 = cis (0)

. The angle stays

0

; only a nonzero modulus would scale — consistent and unsurprising, a useful sanity check.

Does De Moivre work when the modulus

r

is exactly

0

The number is

0

itself, with no defined argument.

0^{n} = 0

for

n > 0

, but the "

n θ

" part is meaningless because there is no angle — the polar form breaks down at the origin.

For

z^{1} = w

, how many "roots" does the formula predict, and is that right?

Exactly one:

k

ranges only over

0

, giving

z_{0} = r cis ϕ = w

. Correct — a first root is just the number itself.

θ

is a negative angle (say

θ = - π /3

), does De Moivre still apply?

Yes. The theorem holds for any real

θ

;

(cis (- π /3))^{n} = cis (- nπ /3)

. Negative angles are simply clockwise arrows and obey the same add-the-angle rule.

For the roots of

z^{n} = w

, what is the angular gap between consecutive roots, regardless of

w

Always

2 π / n

, independent of

ϕ

. Changing

w

's argument rotates the whole set rigidly but never changes the spacing.