Roots of unity — cube roots, nth roots, geometric interpretation: Worked examples

The scenario matrix

Every question about roots of unity (or its cousin $z^{n} = a$ ) falls into one of these cells. The worked examples below each carry a tag like [Cell 3] so you can see the coverage is complete.

#	Case class	What makes it different	Example
1	Plain $z^{n} = 1$ , small $n$	just plug $k = 0 \dots n - 1$	(A)
2	$z^{n} = a$ , $a > 0$ real	scale roots by $r = a^{1/ n}$	(B)
3	$z^{n} = a$ , $a < 0$ real	target angle is $π$ , not $0$	(C)
4	$z^{n} = a$ , $a$ complex (incl. non-QI)	target has a non-trivial angle; quadrant care	(D)
5	Algebra with $ω$ (identities)	reduce exponents mod $n$ , use $1 + ω + ω^{2} = 0$	(E)
6	Sum / product of roots	geometric series, Vieta	(F)
7	Degenerate targets: $n = 1, 2$ and $a = 0$	polygon collapses to a point / segment / origin	(G)
8	Primitive vs non-primitive ( $g cd$ )	which roots regenerate all others	(H)
9	Word problem (real-world)	translate physics into $z^{n} = a$	(I)
10	Exam twist (nested / limiting)	$(1 - ω) (1 - ω^{2}) \dots$ type	(J)

(A) Plain roots — [Cell 1]

Worked example Find all

5th roots of unity and confirm they form a regular pentagon. Forecast: how many points? What angle separates neighbours? (Guess: $5$ points, $7 2^{\circ}$ apart.)

Write the formula $z_{k} = e^{i 2 π k /5}$ for $k = 0, 1, 2, 3, 4$ . Why this step? $z^{5} = 1$ means modulus $r = 1$ and argument $5 θ = 2 π k$ , so $θ = 2 π k /5$ .
Convert each angle to degrees: $0^{\circ}, 7 2^{\circ}, 14 4^{\circ}, 21 6^{\circ}, 28 8^{\circ}$ . Why this step? Degrees make the spacing visible — each step is $36 0^{\circ} /5 = 7 2^{\circ}$ .
Plot them on the Argand Diagram (see figure). Consecutive points are equal chords apart, so they are the vertices of a regular pentagon (a regular polygon). Why this step? Equal angular spacing on a circle $\Rightarrow$ equal chord lengths $\Rightarrow$ regular polygon.

Figure A — Five pink dots at $0^{\circ}, 7 2^{\circ}, 14 4^{\circ}, 21 6^{\circ}, 28 8^{\circ}$ on the unit circle, joined by blue edges into a regular pentagon; yellow spokes from the origin show each is a unit-length vector. Read off that neighbours differ by exactly $7 2^{\circ}$ .

Verify: The five angles sum-check via symmetry: their vector sum should be $0$ . Numerically $\sum_{k} cos (72 k^{\circ}) = 0$ and $\sum_{k} sin (72 k^{\circ}) = 0$ . ✓

(B) $z^{n} = a$ with $a > 0$ real — [Cell 2]

(C) $z^{n} = a$ with $a < 0$ real — [Cell 3]

Worked example Solve

$z^{3} = - 1$ . Forecast: Is $z = - 1$ a solution? Yes — but there must be three. Where are the other two? (Guess: a triangle, one vertex at $- 1$ , rotated so its "start angle" is $π /3$ .)

Write $- 1 = 1 \cdot e^{iπ}$ . Why this step? A negative real number sits at angle $π$ (pointing left), not $0$ — this is the whole point of Cell 3.
Modulus: $r^{3} = 1 \Rightarrow r = 1$ . Why this step? $∣ - 1∣ = 1$ , so all roots are on the unit circle.
Argument: $3 θ = π + 2 π k \Rightarrow θ = \frac{π}{3} + \frac{2 π k}{3}$ , $k = 0, 1, 2$ . Why this step? The target angle $π$ carries into the numerator — this shifts every root by $π /3$ compared with $z^{3} = 1$ .
Evaluate:
- $k = 0 : θ = \frac{π}{3} = 6 0^{\circ} \Rightarrow \frac{1}{2} + \frac{3}{2} i$ .
- $k = 1 : θ = π = 18 0^{\circ} \Rightarrow - 1$ .
- $k = 2 : θ = \frac{5 π}{3} = 30 0^{\circ} \Rightarrow \frac{1}{2} - \frac{3}{2} i$ .
Why this step? Convert each polar form back to $cos θ + i sin θ$ .

Figure C — Solid blue triangle: the three roots of $z^{3} = - 1$ at $6 0^{\circ}, 18 0^{\circ}, 30 0^{\circ}$ (pink dots, yellow spokes). Dashed white triangle: the roots of $z^{3} = 1$ at $0^{\circ}, 12 0^{\circ}, 24 0^{\circ}$ . The solid triangle is the dashed one rotated by $6 0^{\circ}$ — that is what the target angle $π$ does.

Verify: $(\frac{1}{2} + \frac{3}{2} i)^{3} = - 1$ (check below), $(- 1)^{3} = - 1$ . ✓ The triangle for $z^{3} = - 1$ is the $z^{3} = 1$ triangle rotated by $6 0^{\circ}$ . ✓

(D) $z^{n} = a$ with $a$ complex — [Cell 4]

Worked example (D1, QI target) Solve

$z^{2} = 3 + 4 i$ . Forecast: Two answers, negatives of each other. Modulus of each is $∣3 + 4 i ∣ = 5$ . (Guess before finding the angle.)

Modulus of target: $∣3 + 4 i ∣ = 3^{2} + 4^{2} = 25 = 5$ . Why this step? We must know the length before we can halve the angle and root the length.
Root the modulus: $r^{2} = 5 \Rightarrow r = 5$ . Why this step? $∣ z ∣^{2} = ∣ z^{2} ∣$ , so $∣ z ∣ = 5$ .
Argument via $atan2$ : since $3 + 4 i$ is in QI (both parts positive), $θ_{target} = atan2 (4, 3) \approx 53.1 3^{\circ}$ ; solutions have $ϕ = \frac{θ}{2} + 18 0^{\circ} k$ , $k = 0, 1$ . Why this step? Squaring doubles the angle, so halving undoes it — and adding $18 0^{\circ}$ gives the second root. Using $atan2$ (not bare $arctan$ ) guarantees the quadrant is right.
Instead of trig, guess $z = c + d i$ with $c^{2} - d^{2} = 3$ , $2 c d = 4$ . Solving: $c = 2, d = 1$ . So $z = \pm (2 + i)$ . Why this step? For $n = 2$ the "component" method is faster and exact — and dodges all angle-quadrant worries.

Verify: $(2 + i)^{2} = 4 + 4 i + i^{2} = 4 + 4 i - 1 = 3 + 4 i$ . ✓ And $(- 2 - i)^{2} = (2 + i)^{2} = 3 + 4 i$ . ✓

Worked example (D2, non-QI target — why

$atan2$ matters) Solve $z^{2} = - 3 - 4 i$ . Forecast: the target lies in QIII (both parts negative). If you blindly used $arctan (\frac{- 4}{- 3}) = arctan \frac{4}{3} \approx 5 3^{\circ}$ , you'd place the target in QI and get the wrong roots. Where should it really point? (Guess: near $- 12 7^{\circ}$ , i.e. down-left.)

Modulus of target: $∣ - 3 - 4 i ∣ = (- 3)^{2} + (- 4)^{2} = 25 = 5$ . Why this step? Length ignores sign, so it's the same $5$ as in D1 — only the direction differs.
Root the modulus: $r^{2} = 5 \Rightarrow r = 5$ . Why this step? $∣ z ∣ = ∣ z^{2} ∣ = 5$ , same as before.
Correct argument: $atan2 (- 4, - 3) \approx - 126.8 7^{\circ}$ (QIII), not the bare- $arctan$ value $5 3^{\circ}$ . Halving: $ϕ \approx - 63. 4^{\circ}$ or $ϕ + 18 0^{\circ} \approx 116. 6^{\circ}$ . Why this step? This is the whole reason $atan2$ exists — bare $arctan$ cannot tell $- 3 - 4 i$ (QIII) from $3 + 4 i$ (QI), yet their square roots are completely different arrows.
Component method as a clean check: guess $z = c + d i$ with $c^{2} - d^{2} = - 3$ , $2 c d = - 4$ . Solving gives $c = 1, d = - 2$ (and its negative). So $z = \pm (1 - 2 i)$ . Why this step? Exact integers confirm the trig without rounding.

Verify: $(1 - 2 i)^{2} = 1 - 4 i + 4 i^{2} = 1 - 4 i - 4 = - 3 - 4 i$ . ✓ And $(- 1 + 2 i)^{2} = - 3 - 4 i$ . ✓ Note $ar g (- 3 - 4 i)$ is in QIII, so bare $arctan$ would have failed. ✓

(E) Algebra with $ω$ — [Cell 5]

Worked example With

$ω$ a primitive cube root of unity, simplify $(1 - ω) (1 - ω^{2})$ . Forecast: answers to these are almost always small integers. Guess: $3$ ? (We'll confirm, and Cell J proves the general rule $\prod_{k = 1}^{n - 1} (1 - ω^{k}) = n$ .)

Expand: $(1 - ω) (1 - ω^{2}) = 1 - ω - ω^{2} + ω^{3}$ . Why this step? Just multiply out; then we can hit each term with a known identity.
Use $ω^{3} = 1$ : the last term becomes $1$ . Why this step? Reduce exponents mod 3 — $ω^{3}$ is a full loop back to $1$ .
Use $1 + ω + ω^{2} = 0 \Rightarrow - ω - ω^{2} = 1$ . Why this step? The defining cube-root identity collapses the middle two terms.
Combine: $1 + (- ω - ω^{2}) + 1 = 1 + 1 + 1 = 3$ . Why this step? Add the three cleaned-up pieces.

Verify: Substitute $ω = - \frac{1}{2} + \frac{3}{2} i$ numerically — the product is $3$ , matching the general rule $\prod_{k = 1}^{n - 1} (1 - ω^{k}) = n$ with $n = 3$ (proved in Cell J). ✓

(F) Sum and product of roots — [Cell 6]

Worked example For the

6th roots of unity, compute (i) their sum, (ii) their product. Forecast: sum of any $n$ roots ( $n \geq 2$ ) is $0$ . Product of the 6 roots — is it $+ 1$ or $- 1$ ?

Sum: by the Geometric Series argument, $\sum_{k = 0}^{5} ω^{k} = \frac{ω ^{6} - 1}{ω - 1} = \frac{0}{ω - 1} = 0$ . Why this step? $ω^{6} = 1$ kills the numerator; $ω \neq = 1$ keeps the denominator alive.
Product via Vieta on $z^{6} - 1 = 0$ : write the polynomial as $z^{6} + 0 z^{5} + \dots + 0 z + c_{0}$ , where $c_{0}$ is the constant term (here $c_{0} = - 1$ ) and the leading coefficient is $1$ . Vieta's rule says the product of all $n$ roots of a monic degree- $n$ polynomial equals $(- 1)^{n} c_{0}$ . So product $= (- 1)^{6} \cdot (- 1) = - 1$ . Why this step? Multiplying out $\prod (z - z_{k})$ shows the constant term is $(- 1)^{n} \prod z_{k}$ ; matching it to $c_{0}$ gives the product directly, no need for any off-page formula.
Geometric check: the six roots pair into ${1, - 1}$ , ${e^{iπ /3}, e^{- iπ /3}}$ , ${e^{i 2 π /3}, e^{- i 2 π /3}}$ . The pair $1 \cdot (- 1) = - 1$ ; each conjugate pair multiplies to $∣ z ∣^{2} = 1$ . Total $- 1$ . Why this step? Cross-checks Vieta geometrically without leaving this page.

Verify: Numerically $\prod_{k = 0}^{5} e^{i 2 π k /6} = - 1$ and the sum is $0$ . ✓ (For odd $n$ , $c_{0} = - 1$ still but $(- 1)^{n} = - 1$ gives product $+ 1$ ; e.g. $n = 3$ : $1 \cdot ω \cdot ω^{2} = ω^{3} = 1$ .) ✓

(G) Degenerate targets — [Cell 7]

Worked example Handle the degenerate cases: (i)

$z^{1} = 1$ , (ii) $z^{2} = 1$ , (iii) $z^{n} = 0$ . Forecast: for $n = 1$ one root; for $n = 2$ two roots; for target $0$ , one root repeated. No polygon in the usual sense.

$n = 1$ : $z^{1} = 1 \Rightarrow z = 1$ . Just ${1}$ . Why this step? Only $k = 0$ is allowed ( $k = 0 \dots n - 1 = 0$ ). The polygon collapses to a single point.
$n = 2$ : $z_{k} = e^{iπ k}$ , $k = 0, 1 \Rightarrow z = 1, - 1$ . Why this step? Two points $18 0^{\circ}$ apart — the "2-gon" is a diameter (line segment), not an area.
Target $a = 0$ : $z^{n} = 0 \Rightarrow z = 0$ only, a single root of multiplicity $n$ . Why this step? $∣ z ∣^{n} = 0 \Rightarrow r = 0$ , so there's no circle to spread points around — all $n$ roots pile up at the origin. The angle is undefined, which is fine because there is nothing to point.
Sum check: for $n = 1$ , the sum is $1$ (the "sum $= 0$ " rule needs $n \geq 2$ ); for $n = 2$ , $1 + (- 1) = 0$ ; for $z^{n} = 0$ , the sum of roots (all $0$ ) is $0$ too. Why this step? Reminds you the sum-is-zero identity has a boundary case: it fails only for $n = 1$ .

Verify: $1^{1} = 1$ ; $1^{2} = 1$ , $(- 1)^{2} = 1$ ; $0^{n} = 0$ ; and $1 + (- 1) = 0$ but the $n = 1$ sum is $1 \neq = 0$ . ✓

(H) Primitive vs non-primitive — [Cell 8]

Worked example Among the

6th roots $ω^{0}, \dots, ω^{5}$ (where $ω = e^{i 2 π /6}$ ), which are primitive? Forecast: primitivity happens iff $g cd (k, 6) = 1$ . Which $k$ in ${0, 1, 2, 3, 4, 5}$ satisfy that?

Compute $g cd (k, 6)$ for each $k$ : $g cd (0, 6) = 6, g cd (1, 6) = 1, g cd (2, 6) = 2, g cd (3, 6) = 3, g cd (4, 6) = 2, g cd (5, 6) = 1$ . Why this step? By the definition above, a root regenerates all $n$ exactly when $k$ and $n$ share no common factor.
Primitive ones: $k = 1$ and $k = 5$ . Why this step? Only those have $g cd = 1$ .
Sanity: powers of $ω^{1}$ step by $6 0^{\circ}$ each time — hitting all 6 dots. Powers of $ω^{2}$ step by $12 0^{\circ}$ — hitting only ${k = 0, 2, 4}$ (a triangle), so it does not regenerate all six. Why this step? Confirms geometrically why $g cd$ matters: the step size must not close up early.

Verify: The number of primitive 6th roots is $φ (6) = 2$ (Euler's totient), matching ${k = 1, 5}$ . ✓

(I) Word problem — [Cell 9]

Worked example A wheel of radius

$1 m$ has 8 evenly spaced reflectors, one at the 3-o'clock position. Model their positions as complex numbers and find the reflector closest to straight up $(0, 1)$ . Forecast: 8 reflectors = 8th roots of unity. "Straight up" is angle $9 0^{\circ}$ . Is there a reflector exactly there? (Guess: $9 0^{\circ}$ is a multiple of $4 5^{\circ}$ .)

Positions: $z_{k} = e^{i 2 π k /8} = e^{iπ k /4}$ , $k = 0, \dots, 7$ — angles $0^{\circ}, 4 5^{\circ}, 9 0^{\circ}, \dots, 31 5^{\circ}$ . Why this step? 8 equal spacings of $36 0^{\circ} /8 = 4 5^{\circ}$ = 8th roots of unity, radius $1 m$ .
"Straight up" $= 9 0^{\circ}$ , which is $k = 2$ : $z_{2} = e^{iπ /2} = i$ , i.e. the point $(0, 1)$ . Why this step? $9 0^{\circ} = 2 \times 4 5^{\circ}$ , so a reflector sits exactly there.
Answer with units: the reflector at $(0 m, 1 m)$ — exactly at the top. Why this step? Convert the complex number $i$ back to physical coordinates in metres.

Verify: $z_{2} = e^{iπ /2} = i = 0 + 1 i$ , matching $(0, 1)$ . ✓ There are $8$ reflectors and $\sum z_{k} = 0$ (the wheel is balanced). ✓

(J) Exam twist — [Cell 10]

Worked example Prove that

$(1 - ω) (1 - ω^{2}) (1 - ω^{3}) (1 - ω^{4}) = 5$ , where $ω = e^{i 2 π /5}$ is a primitive 5th root of unity. Forecast: a product of " $1$ minus each non-trivial root" of $z^{5} - 1$ . Guess: it equals $n = 5$ . (This is the famous identity Cell E promised.)

Factor $z^{5} - 1 = (z - 1) (z - ω) (z - ω^{2}) (z - ω^{3}) (z - ω^{4})$ . Why this step? Fundamental Theorem of Algebra: the roots are exactly the 5th roots of unity.
Divide by $(z - 1)$ : $\frac{z ^{5} - 1}{z - 1} = 1 + z + z^{2} + z^{3} + z^{4} = (z - ω) (z - ω^{2}) (z - ω^{3}) (z - ω^{4})$ . Why this step? The Geometric Series sum removes the trivial root $z = 1$ , leaving only the non-trivial factors.
Set $z = 1$ : left side $= 1 + 1 + 1 + 1 + 1 = 5$ ; right side $= (1 - ω) (1 - ω^{2}) (1 - ω^{3}) (1 - ω^{4})$ . Why this step? Evaluating the identity at $z = 1$ turns each factor into exactly the product we want.
Therefore the product $= 5$ . In general this trick gives $\prod_{k = 1}^{n - 1} (1 - ω^{k}) = n$ — which is exactly what verified Cell E ( $n = 3 \Rightarrow 3$ ). Why this step? The same argument works for any $n$ : a reusable exam weapon.

Verify: Numerically $\prod_{k = 1}^{4} (1 - e^{i 2 π k /5}) = 5$ , and the same rule at $n = 3$ gives $3$ . ✓

Recall Coverage self-check (cover the answers)

Which cell handles a negative real target? ::: Cell 3 — target angle is $π$ , roots shift by $π / n$ . Why must you use $atan2$ not $arctan$ for a complex target? ::: $arctan$ only returns QI/QIV angles; $atan2$ reads both signs to fix the quadrant (see D2, a QIII target). Why does the sum-is-zero rule need $n \geq 2$ ? ::: For $n = 1$ the only root is $1$ , and $ω - 1 = 0$ breaks the geometric-series step. What are the roots of $z^{n} = 0$ ? ::: A single root $0$ with multiplicity $n$ (all pile at the origin). How do you know which powers of $ω$ are primitive? ::: $g cd (k, n) = 1$ — the generators of the full set (Cell 8). The trick for $\prod (1 - ω^{k})$ ? ::: Factor $\frac{z ^{n} - 1}{z - 1}$ and set $z = 1$ to get $n$ (Cell 10).