nth Roots of Complex Numbers
Solving $z^n = w$ in $\mathbb{C}$ always produces exactly $n$ solutions — and they form a beautifully regular pattern. By writing $w$ in modulus–argument form and applying De Moivre's theorem in reverse, the $n$ roots fall as the vertices of a regular $n$-gon centred on the origin. This lesson turns root extraction into geometry: scale by $R^{1/n}$, divide the angle by $n$, and step around by $2\pi/n$.
Write $1$ in modulus–argument form. Then use De Moivre's theorem to write down $1^{1/3}$ in at least two different ways. Before checking — how many distinct cube roots of $1$ should there be in $\mathbb{C}$? Sketch them on an Argand diagram.
Every $z^n = w$ problem rewards two habits: put $w$ in modulus–argument form first (find $R$ and $\alpha$, including the $+2\pi k$ ambiguity), then divide modulus and argument by $n$ (modulus $\to R^{1/n}$; argument $\to (\alpha + 2\pi k)/n$ for $k = 0, 1, \dots, n-1$). Skipping the $+2\pi k$ step is how students lose all but one root.
The polar-divide-step reading: (1) write $w = R(\cos\alpha + i\sin\alpha)$, (2) the $n$ roots have modulus $R^{1/n}$ and arguments $\dfrac{\alpha + 2\pi k}{n}$, (3) step $k$ from $0$ to $n-1$ to sweep out a regular $n$-gon.
$r_k = R^{1/n}\!\left[\cos\!\dfrac{\alpha + 2\pi k}{n} + i\sin\!\dfrac{\alpha + 2\pi k}{n}\right]$
Key facts
- $z^n = w$ has exactly $n$ distinct complex solutions for $w \neq 0$
- If $w = R(\cos\alpha + i\sin\alpha)$ then $r_k = R^{1/n}\,\text{cis}\,\frac{\alpha + 2\pi k}{n}$, $k = 0,\dots,n-1$
- All roots have modulus $R^{1/n}$
- Consecutive roots differ in argument by $2\pi/n$
Concepts
- Why we need the $+2\pi k$ term (the argument is multi-valued)
- Why the roots sit on a regular $n$-gon centred at the origin
- How De Moivre's theorem connects to root extraction
Skills
- Find all $n$-th roots of a given complex number
- Convert roots between polar and Cartesian form
- Plot all $n$-th roots accurately on an Argand diagram
To solve $z^n = w$ where $w = R(\cos\alpha + i\sin\alpha)$, write $z = r(\cos\theta + i\sin\theta)$ and apply De Moivre's theorem.
Then $z^n = r^n(\cos n\theta + i\sin n\theta) = R(\cos\alpha + i\sin\alpha)$. Equating moduli and arguments (the latter only up to $2\pi$):
- $r^n = R \;\Rightarrow\; r = R^{1/n}$ (the positive real $n$-th root).
- $n\theta = \alpha + 2\pi k$ for some integer $k$, so $\theta = \dfrac{\alpha + 2\pi k}{n}$.
- Stepping $k = 0, 1, 2, \dots, n-1$ produces $n$ distinct roots; $k = n$ recycles back to $k = 0$.
Worked through the hook: $z^3 = 1$. Here $R = 1$, $\alpha = 0$, $n = 3$, so $r = 1$ and $\theta = 2\pi k / 3$.
- $k = 0$: $z_0 = 1$
- $k = 1$: $z_1 = \cos(2\pi/3) + i\sin(2\pi/3) = -\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i$
- $k = 2$: $z_2 = \cos(4\pi/3) + i\sin(4\pi/3) = -\tfrac{1}{2} - \tfrac{\sqrt{3}}{2}i$
Three roots, evenly spaced by $120°$ on the unit circle — the equilateral triangle of cube roots of unity.
Setup: write $w$ in polar form $R\,\text{cis}\,\alpha$ · Formula: $r_k = R^{1/n}\,\text{cis}\,\dfrac{\alpha + 2\pi k}{n}$ for $k = 0, \dots, n-1$ · All roots have the same modulus $R^{1/n}$ · Roots are vertices of a regular $n$-gon, spacing $2\pi/n$
Pause — copy the $n$th root formula $r_k = R^{1/n}\,\text{cis}\,\frac{\alpha+2\pi k}{n}$, the equal-modulus property $R^{1/n}$, and the bracket warning $\frac{\alpha+2\pi k}{n}$ not $\frac{\alpha}{n}+2\pi k$ into your book.
Quick check: How many distinct solutions does the equation $z^5 = 32$ have in $\mathbb{C}$, and what is the common modulus of those solutions?
We just saw that the $n$th roots of $w = R\,\text{cis}\,\alpha$ are $r_k = R^{1/n}\,\text{cis}\,\dfrac{\alpha+2\pi k}{n}$ for $k = 0,\ldots,n-1$, all with the same modulus $R^{1/n}$. That raises a question: these roots all lie at equal angular spacing — do they form a recognisable geometric figure? This card answers it → yes: they are the vertices of a regular $n$-gon inscribed in a circle of radius $R^{1/n}$, with angular spacing $2\pi/n$.
Every $n$-th root of $w$ sits on the circle $|z| = R^{1/n}$. Because consecutive arguments differ by $2\pi/n$, the roots are the vertices of a regular $n$-gon with circumradius $R^{1/n}$, centred at the origin. The "first" vertex sits at angle $\alpha/n$ (the principal root), and the rest are found by rotating around the origin.
- $z^2 = w$ — two roots: diameter through the origin (a "$2$-gon").
- $z^3 = w$ — three roots: equilateral triangle.
- $z^4 = w$ — four roots: square aligned at angle $\alpha/4$.
- $z^n = w$ — regular $n$-gon, side length $2R^{1/n}\sin(\pi/n)$.
All roots lie on a circle of radius $R^{1/n}$ · Roots form a regular $n$-gon, spaced by $2\pi/n$ · Principal root is at angle $\alpha/n$ · Sum of all $n$-th roots of $w$ is $0$ (for $n \geq 2$) · Watch the brackets: $\dfrac{\alpha + 2\pi k}{n}$, not $\dfrac{\alpha}{n} + 2\pi k$
Pause — copy the regular $n$-gon geometry (radius $R^{1/n}$, spacing $2\pi/n$, principal root at angle $\alpha/n$) and the sum-of-all-roots identity ($= 0$ for $n \geq 2$) into your book.
Did you get this? True or false: the six sixth-roots of any non-zero complex number have arguments that differ by $\pi/3$ between consecutive vertices.
Worked examples · 3 in a row, reveal as you go
Find all solutions of $z^3 = i$ in modulus–argument form, then describe their geometric arrangement.
Find both square roots of $-3 + 4i$, expressing your answers in Cartesian form $a + bi$.
Find all four solutions of $z^4 = -16$, expressing each root in Cartesian form. Describe their geometric pattern.
Fill the gap: If $w = R\,\text{cis}\,\alpha$ with $R > 0$, the $n$ solutions of $z^n = w$ all have modulus and consecutive arguments differing by , forming a regular $n$-gon.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the equation $z^4 = 1$ has solutions $1,\;i,\;-1,\;-i$, which form the vertices of a square inscribed in the unit circle.
Activities · practice with the ideas
Find all three solutions of $z^3 = 8$ in Cartesian form. Sketch them on an Argand diagram.
Solve $z^4 = -1$ and write each root in the form $a + bi$.
Find both square roots of $5 - 12i$. Use the algebraic method ($a^2 - b^2 = $ real part, $2ab = $ imaginary part).
Find all five fifth-roots of $32$ and explain geometrically why they sum to zero.
Solve $z^6 = 64$ and explain why exactly two of the six roots are purely real.
Odd one out: Three of these are solutions of $z^4 = 1$. Which one is NOT?
Earlier you guessed how many solutions $z^3 = 1$ has, then predicted $z^3 = -8$.
$z^3 = 1$ has three roots: $1$ and $-\tfrac{1}{2} \pm \tfrac{\sqrt{3}}{2}i$. For $z^3 = -8$ the modulus becomes $8^{1/3} = 2$ and the principal argument is $\pi/3$ (since $-8$ has argument $\pi$). So the roots are $2\,\text{cis}(\pi/3),\;2\,\text{cis}(\pi),\;2\,\text{cis}(5\pi/3) = 1 + \sqrt{3}\,i,\;-2,\;1 - \sqrt{3}\,i$ — still an equilateral triangle, but circumradius doubled to $2$ and rotated $60°$ relative to the cube roots of $1$. Same shape, different scale and orientation. That is the geometric content of $z^n = w$.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find all three cube roots of $-1$ in modulus–argument form. (2 marks)
Q2. Solve $z^3 = 4\sqrt{2}(-1 + i)$, giving each solution in modulus–argument form. (3 marks)
Q3. Let $\omega = \text{cis}(2\pi/5)$. (a) Verify that $\omega$ is a fifth root of $1$. (b) Explain why $1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0$. (c) Hence find $\omega + \omega^2 + \omega^3 + \omega^4$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $z^3 = 8$: $R = 8$, $\alpha = 0$, $r = 2$. Roots $2\,\text{cis}(2\pi k/3)$ for $k = 0, 1, 2$: $z_0 = 2$, $z_1 = -1 + \sqrt{3}\,i$, $z_2 = -1 - \sqrt{3}\,i$. Equilateral triangle on $|z| = 2$.
2. $z^4 = -1$: $R = 1$, $\alpha = \pi$, $r = 1$. Arguments $(2k+1)\pi/4$ for $k = 0,1,2,3$. Roots: $\tfrac{1+i}{\sqrt{2}},\;\tfrac{-1+i}{\sqrt{2}},\;\tfrac{-1-i}{\sqrt{2}},\;\tfrac{1-i}{\sqrt{2}}$.
3. Square roots of $5 - 12i$: $a^2 - b^2 = 5$, $2ab = -12$, $a^2 + b^2 = 13$ (since $|5 - 12i| = 13$). So $a^2 = 9$, $b^2 = 4$, $ab < 0$: roots $3 - 2i$ and $-3 + 2i$.
4. $z^5 = 32$: roots $2\,\text{cis}(2\pi k / 5)$ for $k = 0,\dots,4$. Sum is zero because the vertices of a regular pentagon centred at the origin have centroid $0$.
5. $z^6 = 64$: roots $2\,\text{cis}(\pi k/3)$ for $k = 0,\dots,5$. Real roots correspond to arguments $0$ and $\pi$: $z = 2$ and $z = -2$. The other four are conjugate pairs $\pm 1 \pm \sqrt{3}\,i$.
Q1 (2 marks): $-1 = \text{cis}\,\pi$ [setup]. Roots: $\text{cis}(\pi/3)$, $\text{cis}(\pi) = -1$, $\text{cis}(5\pi/3) = \text{cis}(-\pi/3)$ [1 mark for formula, 1 mark for three correct roots].
Q2 (3 marks): $w = 4\sqrt{2}(-1 + i)$. $|w| = 4\sqrt{2} \cdot \sqrt{2} = 8$ [1]. $\arg w = 3\pi/4$ (second quadrant) [1]. Roots: $z = 8^{1/3}\,\text{cis}\big((3\pi/4 + 2\pi k)/3\big) = 2\,\text{cis}\,\theta_k$ where $\theta_0 = \pi/4$, $\theta_1 = 11\pi/12$, $\theta_2 = -5\pi/12$ (or $19\pi/12$) [1].
Q3 (3 marks): (a) $\omega^5 = \text{cis}(2\pi) = 1$, so $\omega$ is a fifth root of $1$ [1]. (b) $1, \omega, \omega^2, \omega^3, \omega^4$ are all five fifth roots of $1$. Their sum is $0$ since they are vertices of a regular pentagon centred at the origin (or: $z^5 - 1 = (z-1)(z^4 + z^3 + z^2 + z + 1)$; setting $z = \omega \neq 1$ gives the bracketed sum equal to $0$) [1]. (c) From (b), $\omega + \omega^2 + \omega^3 + \omega^4 = -1$ [1].
Five timed questions on extracting $n$-th roots and locating them on the Argand plane. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick $n$-th-root questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.