nth roots of complex numbers — finding all n roots: Exercises

Before we start, two shorthands we use everywhere.

Definition "cis" and the argument convention

$cis α$ means $cos α + i sin α$ — a compact name for "the unit vector pointing at angle $α$ ." So $cis 0 = 1$ (points right) and $cis \frac{π}{2} = i$ (points up).
Choosing $θ = ar g w$ . An angle is only defined up to full turns, so we must fix one convention. We use the principal argument: the unique angle in the range $- π < θ \leq π$ (see Polar Form of Complex Numbers). Concretely: measure the angle from the positive real axis, going anticlockwise for the upper half-plane ( $0$ to $π$ ) and clockwise, i.e. negative, for the lower half-plane ( $0$ down to just above $- π$ ).
This is why $- 2 i$ (pointing straight down) is written $θ = - \frac{π}{2}$ rather than $\frac{3 π}{2}$ . Both name the same direction; they differ by one full turn $2 π$ , so they give identical roots — the RAD-k recipe already sweeps every equivalent angle through the $+ 2 π k$ term. Pick whichever is convenient; just stay consistent within one problem.

Level 1 — Recognition

These test whether you can read off $r$ , $θ$ , $n$ and plug in. No cleverness — just the recipe.

Recall Solution 1.1

WHAT we need: count of roots, and their radius. WHY: A nonzero complex number always has exactly $n$ $n$ -th roots, all on one circle of radius $r^{1/ n}$ .

Here $n = 5$ , so there are ==exactly $5$ == distinct roots. The modulus of $w = 3 + 4 i$ is $r = ∣ w ∣ = 3^{2} + 4^{2} = 25 = 5.$ Common modulus of the roots: $r^{1/5} = 5^{1/5} \approx 1.3797$ .

Answer: $5$ roots, each at distance $5^{1/5} \approx 1.380$ from the origin.

Recall Solution 1.2

WHAT: apply RAD-k with $r = 27$ , $θ = \frac{π}{2}$ , $n = 3$ . Modulus of each root: $2 7^{1/3} = 3$ . WHY divide the angle by $n$ (not root it)? Raising to the $n$ -th power multiplies the angle by $n$ (that is De Moivre's theorem: $(cis ϕ)^{n} = cis n ϕ$ ). To undo a multiplication by $n$ you divide by $n$ — so the root's angle is $θ / n$ , plus the $2 π k$ turns that unlock the other roots. The modulus behaves differently: powering multiplies moduli ( $ρ^{n}$ ), so undoing that needs the $n$ -th root $r^{1/ n}$ . Different operations because angle adds while modulus multiplies. Angle of the $k$ -th root: $ϕ_{k} = \frac{θ + 2 π k}{n} = \frac{\frac{π}{2} + 2 π k}{3} = \frac{π}{6} + \frac{2 π k}{3}, k = 0, 1, 2.$ Answer: $z_{k} = 3 cis (\frac{π}{6} + \frac{2 π k}{3})$ .

Level 2 — Application

Now you compute actual roots and simplify to $a + bi$ form.

Recall Solution 2.1

Step 1 — polar form. $- 2 i$ points straight down with length $2$ , so $r = 2$ and, using the principal-argument convention, $θ = - \frac{π}{2}$ (equivalently $\frac{3 π}{2}$ — same direction, one turn apart, so same roots). Step 2 — RAD-k, $n = 2$ : modulus $2^{1/2} = 2$ ; angles $ϕ_{k} = \frac{- \frac{π}{2} + 2 π k}{2} = - \frac{π}{4} + π k$ .

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

Check: $(1 - i)^{2} = 1 - 2 i + i^{2} = - 2 i .$ ✓ Answer: $z = 1 - i$ and $z = - 1 + i$ (negatives of each other, as square roots always are).

Recall Solution 2.2

Polar: $- 27 = 27 cis π$ , so $r = 27$ , $θ = π$ , $n = 3$ . Modulus of roots: $2 7^{1/3} = 3$ . Angles: $ϕ_{k} = \frac{π + 2 π k}{3}$ .

$k = 0 : ϕ = \frac{π}{3} \Rightarrow 3 cis \frac{π}{3} = 3 (\frac{1}{2} + i \frac{3}{2}) = \frac{3}{2} + i \frac{3 3}{2} .$
$k = 1 : ϕ = π \Rightarrow 3 cis π = - 3.$
$k = 2 : ϕ = \frac{5 π}{3} \Rightarrow 3 cis \frac{5 π}{3} = \frac{3}{2} - i \frac{3 3}{2} .$

Answer: $- 3, \frac{3}{2} \pm i \frac{3 3}{2}$ . The one real root is $- 3$ ; the other two are conjugates.

Level 3 — Analysis

Here you reason about structure: symmetry, sums, which roots lie where.

Recall Solution 3.1

WHAT: a sum and one specific value. WHY sum $= 0$ : the roots are the six vertices of a regular hexagon centred at the origin (Argand Diagram). Opposite vertices are negatives, so the vectors cancel in pairs. Algebraically it is a Geometric Progression: $1 + ω + \dots + ω^{5} = \frac{ω ^{6} - 1}{ω - 1} = \frac{1 - 1}{ω - 1} = 0.$ $ω^{3}$ : using $e^{i θ} = cis θ$ , $ω^{3} = e^{iπ /3 \cdot 3} = e^{iπ} = cis π = - 1.$

Answer: sum $= 0$ ; $ω^{3} = - 1$ .

Recall Solution 3.2

Polar: $16 i = 16 cis \frac{π}{2}$ , $r = 16$ , $θ = \frac{π}{2}$ , $n = 4$ . Modulus: $1 6^{1/4} = 2$ . Angles: $ϕ_{k} = \frac{\frac{π}{2} + 2 π k}{4} = \frac{π}{8} + \frac{π k}{2}$ .

$k = 0 : \frac{π}{8} = 22. 5^{\circ}$ → quadrant I.
$k = 1 : \frac{π}{8} + \frac{π}{2} = \frac{5 π}{8} = 112. 5^{\circ}$ → quadrant II. ✅
$k = 2 : \frac{9 π}{8} = 202. 5^{\circ}$ → quadrant III.
$k = 3 : \frac{13 π}{8} = 292. 5^{\circ}$ → quadrant IV.

Exactly one root sits in quadrant II: $z_{1} = 2 cis \frac{5 π}{8} \approx 2 (cos 112. 5^{\circ} + i sin 112. 5^{\circ}) \approx - 0.7654 + 1.8478 i .$

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

Look at the figure: the four roots march around the circle in exactly $9 0^{\circ}$ steps starting at $22. 5^{\circ}$ , so precisely one lands in each quadrant.

Level 4 — Synthesis

Combine ideas: factor a polynomial with roots, or build a number from a root condition.

Recall Solution 4.1

WHY this works: by the Fundamental Theorem of Algebra, a degree-3 polynomial factors over $C$ as $(z - z_{0}) (z - z_{1}) (z - z_{2})$ . The two complex roots are conjugates, and a conjugate pair multiplies to a real quadratic.

Real factor: $(z - 2)$ . Conjugate pair $z_{1, 2} = - 1 \pm i 3$ . Their quadratic: $(z - (- 1 + i 3)) (z - (- 1 - i 3)) = (z + 1)^{2} - (i 3)^{2} = z^{2} + 2 z + 1 + 3 = z^{2} + 2 z + 4.$ Answer: $z^{3} - 8 = (z - 2) (z^{2} + 2 z + 4)$ . Check: expand $(z - 2) (z^{2} + 2 z + 4) = z^{3} + 2 z^{2} + 4 z - 2 z^{2} - 4 z - 8 = z^{3} - 8.$ ✓

Recall Solution 4.2

Step 1: $z^{4} - z = 0 \Rightarrow z (z^{3} - 1) = 0.$ So either $z = 0$ , or $z^{3} = 1$ . Step 2 — the nonzero part is the cube roots of unity: $z_{k} = cis \frac{2 π k}{3}, k = 0, 1, 2 \Rightarrow 1, - \frac{1}{2} + i \frac{3}{2}, - \frac{1}{2} - i \frac{3}{2} .$ Answer: four solutions — $z = 0, 1, - \frac{1}{2} + i \frac{3}{2}, - \frac{1}{2} - i \frac{3}{2} .$ Notice a degree-4 equation gave 4 solutions, but they did not all sit on one circle — $z = 0$ is the centre. That is fine: this was not a pure $z^{n} = w$ equation with $w \neq = 0$ .

Level 5 — Mastery

Full generality, degenerate cases, and a chained argument.

Recall Solution 5.1

Polar: $- 64 = 64 cis π$ , $r = 64$ , $θ = π$ , $n = 6$ . Modulus: $6 4^{1/6} = 2$ . Angles: $ϕ_{k} = \frac{π + 2 π k}{6} = \frac{π}{6} + \frac{π k}{3}$ , $k = 0, \dots, 5$ . Marching $k$ through its six values:

$k = 0 : ϕ = \frac{π}{6} = 3 0^{\circ} \Rightarrow 2 cis 3 0^{\circ} = 3 + i .$
$k = 1 : ϕ = \frac{π}{2} = 9 0^{\circ} \Rightarrow 2 cis 9 0^{\circ} = 2 i .$
$k = 2 : ϕ = \frac{5 π}{6} = 15 0^{\circ} \Rightarrow 2 cis 15 0^{\circ} = - 3 + i .$
$k = 3 : ϕ = \frac{7 π}{6} = 21 0^{\circ} \Rightarrow 2 cis 21 0^{\circ} = - 3 - i .$
$k = 4 : ϕ = \frac{3 π}{2} = 27 0^{\circ} \Rightarrow 2 cis 27 0^{\circ} = - 2 i .$
$k = 5 : ϕ = \frac{11 π}{6} = 33 0^{\circ} \Rightarrow 2 cis 33 0^{\circ} = 3 - i .$

All six have modulus $2$ and are spaced $\frac{2 π}{6} = 6 0^{\circ}$ apart → a regular hexagon.

Read the figure: the six coloured arrows all reach the same circle of radius $2$ , and each is rotated exactly $6 0^{\circ}$ from its neighbour — starting at $3 0^{\circ}$ ( $k = 0$ , the arrow into the first quadrant) and sweeping anticlockwise. Because $6 0^{\circ} \times 6 = 36 0^{\circ}$ closes the loop with six equal steps, the tips are the six vertices of a regular hexagon. Notice the pairs $k = 0, 3$ ; $k = 1, 4$ ; $k = 2, 5$ point in exactly opposite directions — that opposite-vertex symmetry is why the six roots sum to zero.

Answer: $\pm 3 + i, \pm 2 i, \pm 3 - i$ (six roots).

Recall Solution 5.2

WHAT: solve $z^{n} = 0$ . The only solution is $z = 0$ (a product $z \cdot z \dots z$ equals zero only if a factor is zero). WHY the rule breaks: the formula needs $w = r cis θ$ with $r > 0$ and a defined argument $θ$ . But $0$ has modulus $0$ and no well-defined argument — every direction is equally "toward" the origin. Take $r^{1/ n} = 0$ and every angle collapses to the same point. So instead of $n$ distinct roots you get one root, $z = 0$ , counted with multiplicity $n$ . Answer: $z = 0$ only. The "exactly $n$ distinct roots" statement explicitly requires $w \neq = 0$ .

Recall Solution 5.3

Sum: all five 5th roots of unity sum to $0$ (regular pentagon, symmetry / GP), so $1 + ω + ω^{2} + ω^{3} + ω^{4} = 0.$ Product: add the exponents ( $e^{i θ}$ multiplies by adding angles): $ω \cdot ω^{2} \cdot ω^{3} \cdot ω^{4} = ω^{1 + 2 + 3 + 4} = ω^{10} = (ω^{5})^{2} = 1^{2} = 1.$ Answer: sum $= 0$ , product $= 1$ .

Recall Feynman check: what did every problem share?

Every single solution did the same three things: (1) turn $w$ into a length and an angle, (2) take the ordinary positive root of the length, and (3) fan the angle out into $n$ equal slices. Recognition (L1) tested step 2 vs 3; Analysis (L3) read off where the slices land; Synthesis (L4) fed roots back into polynomials; Mastery (L5) tested the fine print (dividing by zero, $w = 0$ ). The picture never changed: equal length, spread the angle.

Connections

Hinglish version — same content in Hinglish.
De Moivre's Theorem — the engine dividing angles by $n$ .
Roots of Unity — L3 and L5 lean on it.
Euler's Formula — the $e^{i θ} = cis θ$ identity used from L3 on.
Fundamental Theorem of Algebra — guarantees the root count used in L4.
Geometric Progression — why the roots of unity sum to zero.
Polar Form of Complex Numbers — the principal-argument convention.