nth roots of complex numbers — finding all n roots: Worked examples

This is a companion drill page for the parent note. There we built the master formula from scratch. Here we hunt it into every corner — every quadrant of the input angle, the degenerate cases (zero, real, negative real, pure imaginary), a limiting-behaviour case, a word problem, and an exam twist. If you meet a root problem in the wild, it belongs to one of the cells below.

Recall The one formula we keep reusing

If $w = r (cos θ + i sin θ)$ (that is, $r = ∣ w ∣$ is the length and $θ = ar g w$ is the angle), then the $n$ solutions of $z^{n} = w$ are $z_{k} = r^{1/ n} [cos (\frac{θ + 2 π k}{n}) + i sin (\frac{θ + 2 π k}{n})], k = 0, 1, \dots, n - 1.$ Recipe = "Root the length, Add $2 π k$ , Divide by $n$ , run k from $0$ to $n - 1$ ." Built from De Moivre's Theorem on top of Polar Form of Complex Numbers.

Here $cis α$ is just shorthand for $cos α + i sin α$ — a unit arrow pointing at angle $α$ . We write $i$ for the number with $i^{2} = - 1$ , drawn as one step straight up on the Argand Diagram.

The scenario matrix

Every root problem is fixed by two inputs: where the number $w$ sits (its angle/quadrant, or a degenerate position) and which root $n$ we want. The table lists every distinct case class, and the example that nails it.

Cell	Input class $w$	What is tricky	Example
A	Positive real ( $θ = 0$ )	"Real ⇒ one root?" trap	Ex 1 ( $z^{3} = 8$ revisited fully)
B	Pure imaginary, up ( $θ = \frac{π}{2}$ , boundary)	odd $n$ , spread of $120°$	Ex 2 ( $z^{3} = i$ )
B $^{'}$	Pure imaginary, down ( $θ = - \frac{π}{2}$ , boundary)	negative axis angle	Ex 2b ( $z^{2} = - i$ )
C	Negative real ( $θ = π$ )	angle exactly $π$	Ex 3 ( $z^{4} = - 16$ check)
D	Quadrant I input	general non-nice angle	Ex 4 ( $z^{2} = 8 + 8 i 3$ )
E	Quadrant II input (real part negative)	naive $arctan$ gives wrong quadrant	Ex 5 ( $z^{2} = - 1 + i 3$ )
F	Quadrant III input (both parts negative)	naive $arctan$ gives wrong quadrant	Ex 6 ( $z^{3} = - 4 - 4 i 3$ )
G	Quadrant IV input (imag part negative)	negative $arctan$ , add $360°$	Ex 7 ( $z^{2} = 1 - i 3$ )
H	Degenerate $w = 0$	how many roots now?	Ex 8 ( $z^{5} = 0$ )
I	Limiting / large $n$	roots crowd a circle	Ex 9 ( $z^{12} = 1$ , spacing)
J	Word problem (rotation)	translate physics → root	Ex 10 (three equal spins)
K	Exam twist (equation, not $z^{n} = w$ )	rearrange first	Ex 11 ( $z^{4} + z^{2} + 1 = 0$ )

Ex 1 — Cell A: positive real, the "one root?" trap

Worked example

$z^{3} = 8$ , find ALL roots. Forecast: $8$ is a plain real number. Guess how many roots — one, or three? Where do they sit?

Step 1. Write $8$ in polar form: $8 = 8 (cos 0 + i sin 0)$ , so $r = 8$ , $θ = 0$ . Why this step? The formula eats $r$ and $θ$ ; a positive real number lies on the positive $x$ -axis, angle $0$ .

Step 2. Root the length: $r^{1/3} = 8^{1/3} = 2$ . All three roots have length $2$ . Why? Only the modulus gets the ordinary real cube root; the angle is handled separately.

Step 3. Angles: $ϕ_{k} = \frac{0 + 2 π k}{3} = \frac{2 π k}{3}$ for $k = 0, 1, 2$ . Why $k = 0, 1, 2$ ? An $n$ -th root has exactly $n$ distinct angles; here $n = 3$ .

$k = 0 : z_{0} = 2 (cos 0 + i sin 0) = 2.$
$k = 1 : z_{1} = 2 cis \frac{2 π}{3} = 2 (- \frac{1}{2} + i \frac{3}{2}) = - 1 + i 3 .$
$k = 2 : z_{2} = 2 cis \frac{4 π}{3} = - 1 - i 3 .$

Verify: $(- 1 + i 3)^{3}$ : length $2$ cubed $= 8$ , angle $120° \times 3 = 360° = 0°$ , so it equals $8$ . ✓ Three trees planted $120°$ apart on a circle of radius $2$ .

Figure — nth roots of complex numbers — finding all n roots

Ex 2 — Cell B: pure imaginary (up), odd root

Worked example

$z^{3} = i$ . Forecast: $i$ points straight up. Cube roots of an "upward" number — will one of them be real?

Step 1. $i = 1 \cdot cis \frac{π}{2}$ , so $r = 1$ , $θ = \frac{π}{2}$ (that is $90°$ ). Why? $i$ has length $1$ and points at $90°$ on the Argand Diagram.

Step 2. Length of roots: $1^{1/3} = 1$ . They sit on the unit circle.

Step 3. Angles: $ϕ_{k} = \frac{π /2 + 2 π k}{3} = \frac{π}{6} + \frac{2 π k}{3}$ . Why this step? We apply the recipe with $r = 1, θ = \frac{π}{2}, n = 3$ : Add $2 π k$ to the angle and Divide by $3$ , giving $\frac{π}{6} + \frac{2 π k}{3}$ — each step in $k$ turns the root by $\frac{2 π}{3} = 120°$ .

$k = 0 : ϕ = \frac{π}{6} = 30° \Rightarrow z_{0} = cos 30° + i sin 30° = \frac{3}{2} + \frac{1}{2} i .$
$k = 1 : ϕ = \frac{π}{6} + \frac{2 π}{3} = \frac{5 π}{6} = 150° \Rightarrow z_{1} = - \frac{3}{2} + \frac{1}{2} i .$
$k = 2 : ϕ = \frac{π}{6} + \frac{4 π}{3} = \frac{3 π}{2} = 270° \Rightarrow z_{2} = - i .$ Why does $z_{2}$ have zero real part and equal $- i$ ? An angle of $270°$ (equivalently $\frac{3 π}{2}$ rad) points straight down the imaginary axis, where $cos 270° = 0$ and $sin 270° = - 1$ ; a unit arrow there is exactly the number $- i$ .

Verify: $(- i)^{3} = - i^{3} = - (- i) = i$ . ✓ And each step in $k$ adds $120°$ , spacing $\frac{2 π}{3}$ — a regular triangle.

Ex 2b — Cell B $^{'}$ : pure imaginary (down), $θ = - \frac{π}{2}$

Worked example

$z^{2} = - i$ . Forecast: $- i$ points straight down. That axis is a boundary just like $+ i$ , but the angle is negative. Guess where the two square roots land — will they be the same as the roots of $z^{2} = i$ ?

Step 1 — polar form. $- i$ has length $1$ and points along the negative imaginary axis, so $θ = - \frac{π}{2} = - 90°$ (equivalently $\frac{3 π}{2} = 270°$ ; same arrow). Why the negative angle? On the boundary "straight down" is $- 90°$ ; either name works, but we must pick one and keep it — here $- \frac{π}{2}$ keeps the arithmetic light.

Step 2 — root length. $1^{1/2} = 1$ . Both roots sit on the unit circle.

Step 3 — angles. $ϕ_{k} = \frac{- π /2 + 2 π k}{2} = - \frac{π}{4} + π k$ for $k = 0, 1$ . Why this step? Recipe with $r = 1, θ = - \frac{π}{2}, n = 2$ : Add $2 π k$ , Divide by $2$ , giving $- \frac{π}{4} + π k$ — the two square roots sit $180°$ apart, as square roots always do.

$k = 0 : ϕ = - 45° \Rightarrow z_{0} = cos (- 45°) + i sin (- 45°) = \frac{1}{2} (1 - i) .$
$k = 1 : ϕ = 135° \Rightarrow z_{1} = - \frac{1}{2} (1 - i) .$

Verify: $(\frac{1 - i}{2})^{2} = \frac{( 1 - i ) ^{2}}{2} = \frac{1 - 2 i - 1}{2} = - i .$ ✓ Note these are the mirror image (across the real axis) of the roots of $z^{2} = i$ — because $- i$ is the reflection of $i$ .

Ex 3 — Cell C: negative real, angle $= π$

Ex 4 — Cell D: Quadrant I input with a real angle

Worked example

$z^{2} = 8 + 8 i 3$ . Forecast: Both parts positive ⇒ $w$ sits in Quadrant I. Guess its angle before computing.

Step 1 — modulus. $r = 8^{2} + (83)^{2} = 64 + 192 = 256 = 16.$ Why? $∣ w ∣ = x^{2} + y^{2}$ is the arrow's length by Pythagoras.

Step 2 — argument. $tan θ = \frac{8 3}{8} = 3$ . Since both parts are positive we are in Quadrant I, so $θ = \frac{π}{3} = 60°$ (no adjustment needed). Why check the quadrant? $arctan 3$ could be $60°$ or $240°$ ; the positive signs pin it to $60°$ .

Step 3. Root length: $1 6^{1/2} = 4$ ; angles $ϕ_{k} = \frac{π /3 + 2 π k}{2} = \frac{π}{6} + π k$ . Why this step? We now feed $r = 16$ and $θ = \frac{π}{3}$ into the master recipe: Root the length ( $1 6^{1/2} = 4$ , so both roots have length $4$ ), then Add $2 π k$ to the angle and Divide by $n = 2$ , which is exactly $\frac{π /3 + 2 π k}{2} = \frac{π}{6} + π k$ — each step in $k$ turning the root by $π = 180°$ .

$k = 0 : ϕ = \frac{π}{6} = 30° \Rightarrow z_{0} = 4 (cos 30° + i sin 30°) = 23 + 2 i .$
$k = 1 : ϕ = \frac{π}{6} + π = 210° \Rightarrow z_{1} = - 23 - 2 i .$

Verify: $(23 + 2 i)^{2} = 12 + 2 \cdot 23 \cdot 2 i + (2 i)^{2} = 12 + 83 i - 4 = 8 + 83 i .$ ✓ Two square roots, negatives of each other.

Ex 5 — Cell E: Quadrant II, where naive $arctan$ lies

Worked example

$z^{2} = - 1 + i 3$ . Forecast: Real part negative, imaginary part positive ⇒ Quadrant II. Your calculator's $arctan (\frac{3}{- 1})$ returns $- 60°$ — pointing into Quadrant IV. Guess the true angle first.

Step 1 — modulus. $r = (- 1)^{2} + (3)^{2} = 1 + 3 = 2.$ Why? $∣ w ∣ = x^{2} + y^{2}$ , the arrow's length.

Step 2 — argument (careful). $tan θ = \frac{3}{- 1} = - 3$ , base angle $60°$ . But $x < 0, y > 0$ ⇒ Quadrant II, so measure from the negative real axis: $θ = 180° - 60° = 120° = \frac{2 π}{3}$ . Why $180° - 60°$ ? $tan$ repeats every $180°$ and cannot tell Q II from Q IV; the signs ( $x < 0, y > 0$ ) force us into the upper-left, angle $120°$ .

Step 3. Root length $2$ ; angles $ϕ_{k} = \frac{2 π /3 + 2 π k}{2} = \frac{π}{3} + π k$ . Why this step? Feed $r = 2, θ = \frac{2 π}{3}$ into the recipe: Root length $2^{1/2} = 2$ ; Add $2 π k$ and Divide by $n = 2$ ⇒ $\frac{π}{3} + π k$ , each $k$ turning by $180°$ .

$k = 0 : ϕ = 60° \Rightarrow z_{0} = 2 (cos 60° + i sin 60°) = \frac{2}{2} + i \frac{6}{2} .$
$k = 1 : ϕ = 240° \Rightarrow z_{1} = - \frac{2}{2} - i \frac{6}{2} .$

Verify: $z_{0}^{2}$ : length $(2)^{2} = 2$ , angle $60° \times 2 = 120°$ , i.e. $2 cis 120° = 2 (- \frac{1}{2} + i \frac{3}{2}) = - 1 + i 3 .$ ✓

Ex 6 — Cell F: Quadrant III, where naive $arctan$ lies

Worked example

$z^{3} = - 4 - 4 i 3$ . Forecast: Both parts negative ⇒ Quadrant III. If you blindly type $arctan (\frac{- 4 3}{- 4})$ your calculator gives $60°$ — but that points into Quadrant I. Guess the true angle.

Step 1 — modulus. $r = (- 4)^{2} + (- 43)^{2} = 16 + 48 = 64 = 8.$

Step 2 — argument (the careful bit). $tan θ = \frac{- 4 3}{- 4} = 3$ , whose base angle is $60°$ . But $x < 0$ and $y < 0$ ⇒ Quadrant III, so add $180°$ : $θ = 180° + 60° = 240° = \frac{4 π}{3} .$ Why add $180°$ ? $tan$ repeats every $180°$ , so it cannot tell Quadrant I from III. The signs of the real and imaginary parts break the tie — see the two arrows in the figure.

Step 3. Root length: $8^{1/3} = 2$ . Angles: $ϕ_{k} = \frac{4 π /3 + 2 π k}{3} = \frac{4 π}{9} + \frac{2 π k}{3}$ . Why this step? Root length $8^{1/3} = 2$ ; Add $2 π k$ and Divide by $n = 3$ , giving $\frac{4 π}{9} + \frac{2 π k}{3}$ — each $k$ turning by $\frac{2 π}{3} = 120°$ .

$k = 0 : ϕ = \frac{4 π}{9} = 80°.$
$k = 1 : ϕ = 80° + 120° = 200°.$
$k = 2 : ϕ = 80° + 240° = 320°.$ So $z_{k} = 2 cis (80°), 2 cis (200°), 2 cis (320°)$ .

Verify: Take $z_{0} = 2 cis 80°$ : cube ⇒ length $8$ , angle $240°$ , i.e. $8 cis 240° = 8 (- \frac{1}{2} - i \frac{3}{2}) = - 4 - 4 i 3 .$ ✓

Ex 7 — Cell G: Quadrant IV, negative $arctan$

Worked example

$z^{2} = 1 - i 3$ . Forecast: Real part positive, imaginary part negative ⇒ Quadrant IV. Here the naive $arctan$ actually does land in the right quadrant (as a negative angle) — but we want a tidy positive angle. Guess it.

Step 1 — modulus. $r = 1^{2} + (- 3)^{2} = 1 + 3 = 2.$

Step 2 — argument. $tan θ = \frac{- 3}{1} = - 3$ , base angle $60°$ . Since $x > 0, y < 0$ we are in Quadrant IV, so $θ = - 60°$ , or equivalently $θ = 360° - 60° = 300° = \frac{5 π}{3}$ . Why add $360°$ ? $- 60°$ and $300°$ point at the exact same arrow (a full turn apart); we choose the positive $300°$ so the root angles come out positive. Both are fine so long as you stay consistent.

Step 3. Root length $2$ ; angles $ϕ_{k} = \frac{5 π /3 + 2 π k}{2} = \frac{5 π}{6} + π k$ . Why this step? Root $2^{1/2} = 2$ ; Add $2 π k$ , Divide by $n = 2$ ⇒ $\frac{5 π}{6} + π k$ , each $k$ turning by $180°$ .

$k = 0 : ϕ = 150° \Rightarrow z_{0} = 2 cis 150° = - \frac{6}{2} + i \frac{2}{2} .$
$k = 1 : ϕ = 330° \Rightarrow z_{1} = \frac{6}{2} - i \frac{2}{2} .$

Verify: $z_{1}^{2}$ : length $(2)^{2} = 2$ , angle $330° \times 2 = 660° \equiv 300°$ , i.e. $2 cis 300° = 2 (\frac{1}{2} - i \frac{3}{2}) = 1 - i 3 .$ ✓

Ex 8 — Cell H: the degenerate case $w = 0$

Worked example Solve

$z^{5} = 0$ . Forecast: The master formula assumed $w \neq = 0$ . What happens when $w = 0$ — five roots, or something else?

Step 1. $w = 0$ has modulus $r = 0$ and no defined argument (a zero-length arrow points nowhere). Why this breaks the formula: the recipe's "spread the angle" step needs a starting direction $θ = ar g w$ , but $ar g 0$ is undefined — there is no arrow to rotate. The formula simply does not apply.

Step 2. Even if we tried, the root length would be $0^{1/5} = 0$ . Every one of the five "spokes" has length zero, so they all collapse onto the single point at the origin — the regular polygon shrinks to a dot. Why? A polygon of radius $0$ has all its vertices piled at the centre; the angles no longer separate anything.

Step 3 — solve directly instead. $z^{5} = 0 ⟺ z \cdot z \cdot z \cdot z \cdot z = 0 ⟺ z = 0$ . So there is exactly one distinct root, $z = 0$ . Why still call it "five roots"? The Fundamental Theorem of Algebra counts roots with multiplicity: $z^{5} = (z - 0)^{5}$ has the root $0$ repeated $5$ times, so $z = 0$ has multiplicity $5$ . The " $n$ distinct roots" statement of the master formula holds only for $w \neq = 0$ ; the $w = 0$ case is exactly where the $n$ separate points merge into one point of multiplicity $n$ .

Verify: $z^{5} = 0$ factors as $(z - 0)^{5}$ ; the only solution is $z = 0$ , counted $5$ times. ✓ Degenerate: $n$ roots become one point of multiplicity $n$ .

Ex 9 — Cell I: limiting behaviour, roots crowd the circle

Worked example The 12th roots of unity:

$z^{12} = 1$ . How far apart are neighbours, and what happens as $n$ grows? Forecast: With $n = 12$ , is the spacing $30°$ or $12°$ ? And if $n \to \infty$ ?

Step 1. $1 = 1 cis 0$ , so $r = 1, θ = 0$ . Root length $1^{1/12} = 1$ — all on the unit circle. Why this step? Feed $r = 1, θ = 0, n = 12$ into the recipe; the rooted length is $1$ , so every root lands on the unit circle.

Step 2. $z_{k} = cis \frac{2 π k}{12} = cis (30° k)$ , $k = 0, \dots, 11$ . Why this step? Add $2 π k$ (with $θ = 0$ ) and Divide by $12$ gives angle $\frac{2 π k}{12} = 30° k$ . These are the Roots of Unity: powers of $ω = e^{2 π i /12}$ , a Geometric Progression $1, ω, \dots, ω^{11}$ .

Step 3 — spacing. Consecutive roots differ by $\frac{2 π}{12} = \frac{π}{6} = 30°$ . Why this is a limit story: spacing $= \frac{2 π}{n} \to 0$ as $n \to \infty$ ; the $n$ -gon's vertices get denser and the polygon tends to the full circle.

Verify (sum): $\sum_{k = 0}^{11} ω^{k} = \frac{ω ^{12} - 1}{ω - 1} = \frac{1 - 1}{ω - 1} = 0.$ ✓ By symmetry the $12$ arrows cancel exactly.

Ex 10 — Cell J: word problem (equal rotations)

Worked example A robot arm at position

$8$ (i.e. the point $8 + 0 i$ , $8$ units right) is reached by applying the same rotation-and-scaling three times starting from the point $1$ . Each application multiplies the position by a fixed complex number $z$ . Find all possible single moves $z$ . Forecast: "Same move three times, landing on $8$ " — this is secretly $z^{3} = 8$ . How many valid moves?

Step 1 — translate. Three identical steps from $1$ : $1 \cdot z \cdot z \cdot z = z^{3}$ . Requiring the end point $8$ gives $z^{3} = 8$ . Why a root problem? "Repeat a multiplication $n$ times to reach $w$ " is exactly $z^{n} = w$ .

Step 2 — reuse Ex 1. Roots: $z = 2, - 1 + i 3, - 1 - i 3$ . Why? This is the same equation as Ex 1; the three cube roots of $8$ are the three admissible moves. Interpretation: each move scales length by $2$ ( $= 8^{1/3}$ ) and rotates by $0°$ , $120°$ , or $240°$ — all three end at $8$ after three repeats.

Step 3 — physical read. The rotating moves ( $120°, 240°$ ) still land on $8$ because $3 \times 120° = 360° \equiv 0°$ . Why? Three turns of $120°$ complete exactly one full loop, cancelling the rotation and leaving only the length scaling to $8$ .

Verify: $2^{3} = 8$ ; and $(- 1 + i 3)^{3} = 8$ (length $2^{3} = 8$ , angle $3 \times 120° = 360° = 0°$ ). ✓ Three admissible moves.

Ex 11 — Cell K: exam twist (an equation, not $z^{n} = w$ )

Worked example Solve

$z^{4} + z^{2} + 1 = 0$ . Forecast: This is not in the form $z^{n} = w$ . Can we still use roots? Try substituting.

Step 1 — substitute $u = z^{2}$ . Then $u^{2} + u + 1 = 0$ . Why? It turns a quartic into a quadratic we can solve, then peel back.

Step 2 — solve the quadratic. $u = \frac{- 1 \pm 1 - 4}{2} = \frac{- 1 \pm i 3}{2}$ . In polar: both have modulus $1$ , and $u_{+} = cis 120°, u_{-} = cis 240°.$ Why polar? We now need square roots of $u$ , and roots love polar form.

Step 3 — take square roots of each $u$ (that is $z^{2} = u$ ). Why the formula $ϕ = \frac{θ}{2} + 180° k$ ? Taking a square root divides the argument by $2$ (that is the $\frac{θ}{2}$ ), and the recipe's " $+ 2 π k$ before dividing by $n = 2$ " becomes $\frac{2 π k}{2} = π k = 180° k$ — the two square roots are always $180°$ apart because $ar g$ is only fixed up to full turns.

$z^{2} = cis 120°$ : angles $\frac{120°}{2} + 180° k = 60°, 240°$ . So $z = cis 60°, cis 240°$ .
$z^{2} = cis 240°$ : angles $120°, 300°$ . So $z = cis 120°, cis 300°$ .

Step 4 — collect. Four roots: $cis 60°, cis 120°, cis 240°, cis 300°$ — the primitive 6th roots of unity. Why 6th? $z^{4} + z^{2} + 1 = \frac{z ^{6} - 1}{z ^{2} - 1}$ , so its roots are 6th-roots of unity that are not square-roots of unity.

Verify: For $z = cis 60°$ : $z^{2} = cis 120°$ , $z^{4} = cis 240°$ , and $cis 240° + cis 120° + 1 = (- \frac{1}{2} - i \frac{3}{2}) + (- \frac{1}{2} + i \frac{3}{2}) + 1 = 0.$ ✓

Recall One-line summary of the matrix

Every case reduces to: find $r = ∣ w ∣$ and $θ = ar g w$ with the correct quadrant, then run "Root, Add $2 π k$ , Divide, k=0..n−1." Degenerate $w = 0$ gives a single root of multiplicity $n$ ; equations not of the form $z^{n} = w$ get a substitution first.

Which two facts fully determine a root problem's answer?

The polar data of

w

(

r

and correctly-quadranted

θ

) and the index

n

What is

ar g 0

, and how many distinct roots does

z^{n} = 0

have?

Undefined; exactly one distinct root

z = 0

, of multiplicity

n

Why must you subtract

60°

from

180°

for a Quadrant II input?

Because

tan

has period

180°

; the signs (

x < 0, y > 0

) put the arrow upper-left, so

θ = 180° - base angle

For a Quadrant IV input why can you add

360°

to the raw negative

arctan

A full turn names the same direction;

300°

and

- 60°

are the same arrow, and

+ 300°

keeps root angles positive.

For an equation like

z^{4} + z^{2} + 1 = 0

, first move?

Substitute

u = z^{2}

to get a quadratic, solve, then take square roots.

Connections

nth roots of complex numbers — finding all n roots — parent note with the master derivation.
De Moivre's Theorem · Polar Form of Complex Numbers · Euler's Formula — the machinery.
Roots of Unity · Geometric Progression — Ex 9 & 11.
Fundamental Theorem of Algebra — why " $n$ roots" (Ex 1, 8).
Argand Diagram — the picture behind every example.