Visual walkthrough — nth roots of complex numbers — finding all n roots

Step 1 — A complex number is an arrow on a grid

WHAT. Draw two number lines crossing at a point $O$ (the origin). Horizontal = ordinary "real" numbers. Vertical = a new direction we call "imaginary", counted in units of $i$. A complex number like $3+2i$ is then the single arrow from $O$ to the point $3$ steps right, $2$ steps up. This picture is an Argand diagram.

WHY. We want to see multiplication and roots, and you cannot see algebra. Turning a number into an arrow lets every operation become a motion (stretch, spin) we can watch.

PICTURE. The red arrow below is the number $w$. Two numbers describe it completely: how long it is, and which way it points.

Step 2 — Rename the arrow by length-and-angle (polar form)

WHAT. Instead of "right/up" coordinates, describe the same arrow by (length, angle). If the length is $r$ and the angle is $\theta$, then trigonometry of the right triangle gives the horizontal part $r\cos \theta$ and the vertical part $r\sin \theta$, so $w=\underset{\text{length}}{\underbrace{r}}(\underset{\text{how far right}}{\underbrace{\cos \theta }}+i\underset{\text{how far up}}{\underbrace{\sin \theta }}).$

WHY. Because — and this is the whole reason polar form exists — multiplying two arrows multiplies their lengths and adds their angles. Roots are un-multiplying, so we want the language where multiplication is simplest. See Polar Form of Complex Numbers.

PICTURE. The dashed legs are $r\cos \theta$ (along the floor) and $r\sin \theta$ (up the wall); the angle $\theta$ sits at the origin.

Step 3 — Powering an arrow: spin faster, stretch harder

WHAT. Take an unknown arrow and call its length $\rho$ (Greek letter "rho") and its angle $\varphi$ ("phi"), so $z=\rho \text{cis}\varphi$. Here $\rho =|z|$ is the modulus of the root (its length, always $\ge 0$) and $\varphi =\arg z$ is the argument of the root (its angle) — the two unknowns we are hunting for. Raise it to the power $n$; De Moivre's Theorem says the length gets powered and the angle gets multiplied by $n$: $z^n=\underset{\text{length powered}}{\underbrace{{\rho }^n}}\text{cis}\underset{\text{angle}\times n}{\underbrace{(n\varphi )}}.$

WHY. Because $z^n$ means multiplying $z$ by itself $n$ times, and each multiplication adds $\varphi$ to the angle ($\varphi +\varphi +\cdots =n\varphi$) and multiplies the length ($\rho \cdot \rho \cdots ={\rho }^n$). Powering is just Step 2's rule applied repeatedly.

PICTURE. Watch the blue arrow of length $\rho =1.15$ at $\varphi =40^{\circ }$ become, after cubing, a longer arrow at $120^{\circ }$. The angle tripled; the length grew.

Step 4 — Reverse the machine: what must a root satisfy?

WHAT. We want $z$ with $z^n=w$. Plug in polar forms for both sides and demand they be the same arrow: ${\rho }^n\text{cis}(n\varphi )=r\text{cis}\theta .$ Two arrows are identical exactly when their lengths match and their angles point the same way: $\underset{\text{lengths}}{\underbrace{{\rho }^n=r}},\underset{\text{angles}}{\underbrace{n\varphi =\theta }}(\text{so far}).$

WHY. Powering multiplied the angle by $n$; to undo it we must divide by $n$. Powering powered the length; to undo it we must take the $n$-th root of the length. Root = reverse of power, term by term.

PICTURE. Left arrow is the target $w$; right arrow is the mystery root $z$ whose triple-spin must land on $w$.

Step 5 — The secret: one arrow wears many angles

WHAT. Here is the move the whole formula turns on. The arrow $w$ points at angle $\theta$ — but also at $\theta +2\pi$, at $\theta +4\pi$, at $\theta +2\pi k$ for any whole number $k$. Spinning a full turn ($2\pi$ radians $=360^{\circ }$) brings you back to the identical arrow. (This is precisely the "$\arg$ is multivalued" fact from Step 1, now put to work.)

So the honest angle equation is not $n\varphi =\theta$ but $n\varphi =\theta +2\pi k,k=\dots ,-1,0,1,2,\dots$

WHY. $\cos$ and $\sin$ repeat every $2\pi$. The picture of $w$ cannot tell $\theta$ from $\theta +2\pi$ — same point, same everything. If we forget the $+2\pi k$ we throw away every root but one.

PICTURE. All three labels $\theta$, $\theta +2\pi$, $\theta +4\pi$ name the same red arrow — the extra loops are drawn spiralling out just so you can see them, but they land on the identical direction.

Step 6 — Solve for length and angle

WHAT. Read off the two equations from Steps 4–5: ${\rho }^n=r⟹\rho =\underset{\text{ordinary positive }n\text{-th root}}{\underbrace{r^{1/n}}},$ $n\varphi =\theta +2\pi k⟹{\varphi }_k=\frac{\theta +2\pi k}{n}.$

WHY each piece.

• $\rho =r^{1/n}$: the length is the positive real $n$-th root of the modulus only. We never root the angle and never root a negative — $r=|w|\ge 0$ always.
• $\frac{\theta +2\pi k}{n}$: add the full turns first, then divide by $n$. Dividing is what pulls the many-angles apart.

PICTURE. The number line of allowed $k$ feeds into the angle formula; each $k$ produces one candidate direction.

Step 7 — Why exactly $n$ roots, not infinitely many

WHAT. We let $k$ run over all integers — surely that gives infinitely many roots? No. Compare $k$ and $k+n$: ${\varphi }_{k+n}=\frac{\theta +2\pi (k+n)}{n}=\frac{\theta +2\pi k}{n}+\underset{=2\pi }{\underbrace{\frac{2\pi n}{n}}}={\varphi }_k+2\pi .$ Adding $2\pi$ is a full turn — same arrow. So the roots cycle with period $n$. Choosing $k=0,1,2,\dots ,n-1$ collects each distinct one exactly once.

WHY. The one extra full turn we injected in Step 5, divided by $n$, reappears as a full turn only after $n$ steps. Between those steps, each ${\varphi }_k$ advances by a fraction $\frac{2\pi }{n}$ of a turn — a genuinely new direction each time.

PICTURE. Consecutive roots sit $\frac{2\pi }{n}$ apart; after $n$ hops you are back at the start. That even spacing is why the roots are the corners of a regular $n$-gon (see Roots of Unity).

Step 8 — Edge & degenerate cases (never get ambushed)

WHAT / WHY / PICTURE, one panel each below.

• $w=0$. Then $r=0$, so $\rho =0^{1/n}=0$. Every root is the single point $z=0$ — the regular polygon collapses to its centre. There is only one root, of "multiplicity $n$". This is the only time the count is not $n$.
• $w$ real and positive (like $8$): $\theta =0$. One root sits on the positive real axis; the rest fan out. Real value does not mean one root — the cubic $z^3=8$ still has $3$.
• $w$ real and negative (like $-16$): $\theta =\pi$. No root lies on the positive real axis; they straddle it symmetrically. This is why real-coefficient equations give conjugate pairs.
• $w$ purely imaginary (like $i$, so $z^2=i$): the arrow points straight up the imaginary axis, so $\theta =\frac{\pi }{2}$ (or straight down, $-\frac{\pi }{2}$, if $w=-i$). Nothing special breaks — you feed $\theta =\frac{\pi }{2}$ into the same formula. For $z^2=i$ the two roots come out at $\frac{\pi }{4}$ and $\frac{5\pi }{4}$: an opposite pair tilted $45^{\circ }$, matching the bottom-right panel below.
• $n=2$ (square roots): the two roots are $\frac{2\pi }{2}=\pi$ apart — exact opposites, $z_1=-z_0$.

PICTURE. Four mini-Argands: the collapsed point ($w=0$), and the $n$-gons for $z^3=8$, $z^4=-16$, and the purely-imaginary $z^2=i$, so every sign and the zero case are on screen at once.

The one-picture summary

Everything on one board: a target arrow $w=r\text{cis}\theta$, its length collapsed to $r^{1/n}$, its angle $\theta$ carrying $+2\pi k$ full turns before being divided by $n$, and the resulting evenly-spaced regular $n$-gon of roots.

WHY this one picture is the whole formula. Read the diagram left-to-right as a recipe. The pink target arrow supplies the two raw ingredients, length $r$ and angle $\theta$. The circle's radius has shrunk to $r^{1/n}$ — that is the length-root, Step 6's first equation, drawn. The yellow root arrows all sit on that circle (same length) but fan out because $\theta$ was first loaded with full turns $\theta +2\pi k$ and then divided by $n$ — Step 5 plus Step 6's second equation, drawn. The equal $\frac{2\pi }{n}$ gaps closing into the blue regular $n$-gon are Step 7, drawn. So each visual feature is one algebraic move: length-root, add-turns-then-divide, even spacing.

Active Recall

Connections

• Parent topic
• De Moivre's Theorem · Polar Form of Complex Numbers · Argand Diagram
• Euler's Formula · Roots of Unity · Geometric Progression · Fundamental Theorem of Algebra