nth roots of complex numbers — finding all n roots

WHAT are we solving?

The special case $w=1$ gives the $n$ -th roots of unity (solutions of $z^n=1$ ).

HOW to derive the formula (from scratch)

Step 1 — Write $w$ in polar form. Any nonzero $w = r(\cos\theta + i\sin\theta)$ , where $r=|w|>0$ and $\theta=\arg w$ . Why this step? Multiplication of complex numbers is easiest in polar form: moduli multiply, arguments add. Roots are about undoing multiplication, so polar is the natural language.

Step 2 — Write the unknown root in polar form too. Let $z = \rho(\cos\phi + i\sin\phi)$ with unknown modulus $\rho>0$ and unknown angle $\phi$ .

Step 3 — Apply De Moivre to $z^n$ . $z^n = \rho^n\big(\cos n\phi + i\sin n\phi\big).$ Why this step? De Moivre's theorem $(\cos\phi+i\sin\phi)^n = \cos n\phi + i\sin n\phi$ tells us powering multiplies the angle by $n$ and powers the modulus.

Step 4 — Match with $w$ . We need $z^n=w$ , so $\rho^n(\cos n\phi + i\sin n\phi) = r(\cos\theta+i\sin\theta).$ Two complex numbers in polar form are equal iff their moduli are equal and their arguments differ by a multiple of $2\pi$ : $\rho^n = r, \qquad n\phi = \theta + 2\pi k,\quad k\in\mathbb{Z}.$ Why the $+2\pi k$ ? THIS is the crucial move. Angles are only defined up to full turns; $\cos$ and $\sin$ have period $2\pi$ . If we forget the $2\pi k$ we get only one root. Keeping it unlocks all $n$ .

Step 5 — Solve. $\rho = r^{1/n}\ (\text{real positive $n$-th root}), \qquad \phi = \frac{\theta + 2\pi k}{n}.$

Step 6 — Find how many distinct roots. As $k=0,1,2,\dots,n-1$ we get $n$ different angles. For $k=n$ the angle is $\frac{\theta}{n}+2\pi$ — same direction as $k=0$ . So values repeat with period $n$ : only $n$ distinct roots.

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

Roots of unity (special case $w=1$ )

Here $r=1$ , $\theta=0$ , so $z_k = \cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n} = e^{2\pi i k/n},\quad k=0,\dots,n-1.$ Writing $\omega = e^{2\pi i/n}$ , the roots are $1, \omega, \omega^2, \dots, \omega^{n-1}$ — a geometric progression.

Worked Examples

Worked example Example 1 — Square roots of

$i$ Solve $z^2 = i$ . Polar form: $i = 1\cdot(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2})$ , so $r=1,\ \theta=\frac{\pi}{2}$ . Why? We need modulus and argument to feed the formula.

Modulus of roots: $r^{1/2}=1$ . Angles: $\phi_k = \dfrac{\pi/2 + 2\pi k}{2}=\dfrac{\pi}{4}+\pi k$ .

$k=0:\ \phi=\frac{\pi}{4}\Rightarrow z_0=\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}=\frac{1}{\sqrt2}(1+i).$
$k=1:\ \phi=\frac{5\pi}{4}\Rightarrow z_1=-\frac{1}{\sqrt2}(1+i).$

Check: $\left(\frac{1+i}{\sqrt2}\right)^2 = \frac{(1+i)^2}{2}=\frac{2i}{2}=i.$ ✓ The two roots are negatives of each other (as expected — square roots always are).

Worked example Example 2 — Cube roots of

$8$ Solve $z^3 = 8$ . Polar: $8=8(\cos 0 + i\sin 0)$ , $r=8,\ \theta=0$ . Modulus of roots: $8^{1/3}=2$ . Angles: $\phi_k = \dfrac{0+2\pi k}{3}=\dfrac{2\pi k}{3}$ .

$k=0:\ z_0 = 2(\cos 0 + i\sin 0)=2.$
$k=1:\ z_1 = 2(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}) = 2(-\tfrac12 + i\tfrac{\sqrt3}{2}) = -1+i\sqrt3.$
$k=2:\ z_2 = 2(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}) = -1-i\sqrt3.$

Why three roots for a real number? Even though $8$ is real, a cubic $z^3-8=0$ has 3 roots; only one is real, the other two are complex conjugates. Real answers ≠ only one root.

Worked example Example 3 — All 4th roots of

$-16$ Solve $z^4 = -16$ . Polar: $-16 = 16(\cos\pi + i\sin\pi)$ , $r=16,\ \theta=\pi$ . Modulus: $16^{1/4}=2$ . Angles: $\phi_k=\dfrac{\pi+2\pi k}{4}$ for $k=0,1,2,3$ : $\frac{\pi}{4},\ \frac{3\pi}{4},\ \frac{5\pi}{4},\ \frac{7\pi}{4}$ . Roots: $2\,\text{cis}\,\frac{\pi}{4}=\sqrt2+i\sqrt2$ , then $-\sqrt2+i\sqrt2$ , $-\sqrt2-i\sqrt2$ , $\sqrt2-i\sqrt2$ . A perfect square rotated $45°$ , radius $2$ . Why evenly spaced? Each step in $k$ adds $\frac{2\pi}{4}=90°$ .

Common Mistakes

Recall Feynman: explain to a 12-year-old

Imagine a clock hand. "Powering to $n$ " means spinning the hand $n$ times faster. To undo it (take the root) you must spin $n$ times slower. But here's the trick: the hand pointing at 12 o'clock is the same as pointing at 12 after going all the way around once, twice, three times... When you slow all those "same" positions down by $n$ , they spread out into $n$ different spots evenly around the clock face — like planting $n$ trees equally around a circular pond. All the trees are the same distance from the centre (that distance is the slow-down of the length), just at different angles.

Active Recall Flashcards

#flashcards/maths

How many distinct

n

-th roots does a nonzero complex number have?

Exactly

n

Formula for the

k

-th root of

z^n=w

where

w=r\text{cis}\,\theta

z_k=r^{1/n}\,\text{cis}\!\left(\dfrac{\theta+2\pi k}{n}\right)

k=0,\dots,n-1

What is the common modulus of all

n

roots?

r^{1/n}

(real positive

n

-th root of

|w|

Angular separation between consecutive roots?

\dfrac{2\pi}{n}

radians.

Geometric shape formed by the

n

roots in the Argand plane?

Vertices of a regular

n

-gon on a circle of radius

r^{1/n}

Why must we add

2\pi k

before dividing by

n

Because

\arg

is defined only mod

2\pi

; adding full turns produces the other distinct roots.

Sum of all

n

-th roots of unity (

n\ge2

0

(symmetry / GP sum).

The

n

-th roots of unity as a GP?

1,\omega,\omega^2,\dots,\omega^{n-1}

with

\omega=e^{2\pi i/n}

Square roots of any complex number relate how?

They are negatives of each other,

z_1=-z_0

For

z^3=8

, why are there complex roots though

8

is real?

A cubic over

\mathbb{C}

has 3 roots; real value ≠ single root.

Connections

De Moivre's Theorem — engine behind angle multiplication/division.
Polar Form of Complex Numbers — prerequisite representation.
Roots of Unity — special case $w=1$ ; cyclotomic structure.
Argand Diagram — geometric visualisation as regular polygons.
Euler's Formula — $e^{i\theta}$ makes the GP structure obvious.
Fundamental Theorem of Algebra — why $z^n-w$ has exactly $n$ roots.
Geometric Progression — sum of roots of unity.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, complex number ka $n$ -th root nikalna matlab wo saare $z$ dhundhna jinke liye $z^n = w$ hota hai. Sabse pehle $w$ ko polar form me likho: $w = r\,\text{cis}\,\theta$ , jahan $r=|w|$ aur $\theta=\arg w$ . Ab De Moivre kehta hai ki power lene se angle $n$ se multiply hota hai — to root lene ke liye angle ko $n$ se divide karna padega, aur modulus ka simple positive $n$ -th root $r^{1/n}$ lena hai.

Asli twist yahan hai: angle to $2\pi$ ke multiples add karke bhi wahi number deta hai (ghadi ki sui poora ghoom ke wapas wahi jagah). Isliye hum likhte hain angle $=\dfrac{\theta+2\pi k}{n}$ , aur $k=0,1,\dots,n-1$ daalte hain. Har $k$ ek alag root deta hai — total $n$ roots. Agar tum $2\pi k$ bhool gaye to sirf ek hi root milega, jo galat hai.

Geometrically ye saare roots ek hi circle par baithte hain jiska radius $r^{1/n}$ hai, aur do consecutive roots ke beech ka angle $\frac{2\pi}{n}$ hota hai — matlab ek perfect regular polygon ban jaata hai Argand plane par. Roots of unity ( $z^n=1$ ) me to ye aur bhi sundar hai: saare roots milke $0$ ban jaate hain symmetry ki wajah se.

Yaad rakhne ka easy tareeka: "Same length, spread the angle" — length ka root lo, angle ko phaila do. Exam me steps clean rakho: polar form → modulus ka root → $\frac{\theta+2\pi k}{n}$ → $k=0$ se $n-1$ tak. Bas ho gaya!

nth roots of complex numbers — finding all n roots

WHAT are we solving?

HOW to derive the formula (from scratch)

Roots of unity (special case $w=1$ )

Worked Examples

Common Mistakes

Active Recall Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Complex Numbers

Connections