Visual walkthrough — De Moivre's theorem — statement, proof, applications

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Step 1 — What a complex number looks like

WHAT. A complex number $z=x+iy$ is just an instruction: "go $x$ to the right, then $y$ upward." That lands you at a single point on a flat sheet. This sheet is called the Argand diagram — think ordinary graph paper, but the horizontal axis is labelled "real" and the vertical axis is labelled "imaginary".

WHY draw it as an arrow. Because we are about to multiply these numbers, and multiplication is going to turn out to be a motion (a spin + a stretch). Motions are easiest to see on arrows, so we draw $z$ as an arrow from the origin $O$ to the point $(x,y)$.

PICTURE. Look at the figure. The green arrow ends at the dot $z=x+iy$. Two numbers fully describe that arrow:

• its length $r$ — how far from the origin, measured by Pythagoras $r=\sqrt{x^2+y^2}$;
• its angle $\theta$ — how far it has swung anticlockwise from the positive real axis.

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Step 2 — Rebuilding $z$ from length and angle

WHAT. If I only tell you the length $r$ and angle $\theta$, can you get back $x$ and $y$? Yes. Drop a vertical line from the arrow-tip to the real axis. You have made a right triangle whose slanted side (hypotenuse) is the arrow of length $r$.

WHY these two ratios. In a right triangle:

• the side along the axis (adjacent to $\theta$) is $r\cos \theta$, because $\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{x}{r}$;
• the side going up (opposite $\theta$) is $r\sin \theta$, because $\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}=\frac{y}{r}$.

$\cos$ and $\sin$ are exactly the tools that convert "an angle" into "how much horizontal / how much vertical" — that is the whole reason they show up here.

PICTURE. The triangle below has its horizontal leg painted orange ($x=r\cos \theta$) and its vertical leg painted magenta ($y=r\sin \theta$).

$z=\underset{\text{length}}{\underbrace{r}}(\underset{\text{horizontal fraction}}{\underbrace{\cos \theta }}+i\underset{\text{vertical fraction}}{\underbrace{\sin \theta }})$

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Step 3 — The one fact everything rests on: multiplying two arrows

WHAT. Take two unit arrows, $z_1=\mathrm{cis}\alpha$ and $z_2=\mathrm{cis}\beta$. Multiply them like ordinary brackets: $z_1{z}_2=(\cos \alpha +i\sin \alpha )(\cos \beta +i\sin \beta ).$

WHY expand. To separate the answer into its horizontal part (real) and vertical part (imaginary), so we can read off where the new arrow points. Multiply the four pairs, and use $i^2=-1$: $z_1{z}_2=\underset{\text{real part}}{\underbrace{(\cos \alpha \cos \beta -\sin \alpha \sin \beta )}}+i\underset{\text{imaginary part}}{\underbrace{(\sin \alpha \cos \beta +\cos \alpha \sin \beta )}}.$

Now look hard at those two brackets. They are exactly the angle-addition formulas: $\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta ,\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta .$ So the whole product collapses to $z_1{z}_2=\cos (\alpha +\beta )+i\sin (\alpha +\beta )=\mathrm{cis}(\alpha +\beta ).$

PICTURE. The blue arrow sits at angle $\alpha$, the violet arrow at angle $\beta$; their product (magenta) sits at angle $\alpha +\beta$. Multiplying added the two angles. (Lengths would multiply too — here both are $1$, so $1\times 1=1$.)

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Step 4 — Powers of a positive integer: spin $n$ times

WHAT. What is $(\mathrm{cis}\theta {)}^n$? It means multiplying $\mathrm{cis}\theta$ by itself $n$ times.

WHY it's easy now. Each multiplication adds one more $\theta$ (Step 3). Start at angle $\theta$; after squaring you are at $\theta +\theta =2\theta$; after cubing, $3\theta$; after $n$ multiplications, $n\theta$.

Formally this is a proof by induction:

• Base $n=1$: $(\mathrm{cis}\theta {)}^1=\mathrm{cis}(1\cdot \theta )$. ✔
• Step: if $(\mathrm{cis}\theta {)}^k=\mathrm{cis}(k\theta )$, then $(\mathrm{cis}\theta {)}^{k+1}=\underset{\text{assumed}}{\underbrace{\mathrm{cis}(k\theta )}}\cdot \underset{\text{one more}}{\underbrace{\mathrm{cis}\theta }}=\mathrm{cis}(k\theta +\theta )=\mathrm{cis}((k+1)\theta ).$ The middle equality is exactly the master fact from Step 3 — that is the only tool we needed.

PICTURE. Watch the arrow march around the circle: $\theta ,2\theta ,3\theta ,\dots$, each step the same swing $\theta$, like a ratchet.

For a general length $r$, each multiplication also multiplies the length by $r$, so: $(r\mathrm{cis}\theta {)}^n=\underset{{\text{length}}^n}{\underbrace{r^n}}\mathrm{cis}\underset{\text{angle}\times n}{\underbrace{(n\theta )}}.$

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Step 5 — The degenerate case $n=0$

WHAT. Anything (non-zero) to the power $0$ is $1$.

WHY it still fits. $\mathrm{cis}(0\cdot \theta )=\mathrm{cis}0=\cos 0+i\sin 0=1+0i=1$. So the formula gives $1$ — exactly the arrow of length $1$ pointing straight right, having swung no angle. The pattern doesn't break at zero; it just means "don't spin at all."

PICTURE. A single arrow lying flat on the positive real axis, angle $0$.

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Step 6 — Negative powers: spin backwards

WHAT. What about $(\mathrm{cis}\theta {)}^{-m}$ for $m>0$? A negative exponent means reciprocal: $\frac{1}{(\mathrm{cis}\theta {)}^m}=\frac{1}{\mathrm{cis}(m\theta )}$.

WHY the reciprocal is a clean spin-backwards. The arrow $\mathrm{cis}(m\theta )$ has length $1$, so ${\cos }^2(m\theta )+{\sin }^2(m\theta )=1$. Multiply top and bottom by the conjugate: $\frac{1}{\cos m\theta +i\sin m\theta }=\frac{\cos m\theta -i\sin m\theta }{\underset{=1}{\underbrace{{\cos }^{2}m\theta +{\sin }^{2}m\theta }}}=\cos m\theta -i\sin m\theta .$ And $\cos m\theta -i\sin m\theta =\cos (-m\theta )+i\sin (-m\theta )=\mathrm{dcis}(-m\theta )$, because $\cos$ is unchanged by a sign flip while $\sin$ flips sign. So $(\mathrm{cis}\theta {)}^{-m}=\mathrm{cis}(-m\theta ).$ Taking a negative power just spins the arrow the other way (clockwise). The theorem now holds for every integer $n$: positive, zero, negative.

PICTURE. The forward arrow at $+m\theta$ (magenta) and its reciprocal at $-m\theta$ (violet), mirror images across the real axis.

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Step 7 — Reversing the machine: $n$-th roots

WHAT. Solve $z^n=w$, where $w=r\mathrm{cis}\varphi$. Write the unknown as $z=\rho \mathrm{cis}\psi$. Then $z^n={\rho }^n\mathrm{cis}(n\psi )$ must equal $r\mathrm{cis}\varphi$.

WHY the $2\pi k$ appears. Matching lengths gives $\rho =r^{1/n}$ (only one positive answer). Matching angles gives $n\psi =\varphi$ — but an arrow at angle $\varphi$ is the same arrow as at $\varphi +2\pi$, $\varphi +4\pi$, …, since a full turn ($2\pi$) lands you home. Every one of those is a legal target: $n\psi =\varphi +2\pi k⟹\psi =\frac{\varphi +2\pi k}{n}.$ As $k=0,1,\dots ,n-1$ we get $n$ distinct angles, each $\frac{2\pi }{n}$ apart. At $k=n$ we've added a whole $2\pi$ back and repeat, so there are exactly $n$ roots.

$\boxed{z_k=r^{1/n}\mathrm{cis}\left(\frac{\varphi +2\pi k}{n}\right),k=0,1,\dots ,n-1.}$

PICTURE. The $n$ roots sit like clock numbers, evenly spaced on a circle of radius $r^{1/n}$ — here the cube roots of unity (see Roots of unity), an equilateral triangle.

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Step 8 — The other payoff: multiple-angle identities

WHAT. Expand $(\cos \theta +i\sin \theta {)}^n$ with the Binomial theorem, then match real and imaginary parts against $\cos n\theta +i\sin n\theta$.

WHY it works. De Moivre says these two expressions are the same complex number; two complex numbers are equal only if their real parts match AND their imaginary parts match. For $n=3$, with $(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$, $a=\cos \theta$, $b=i\sin \theta$, and $i^2=-1,i^3=-i$: $(\cos \theta +i\sin \theta {)}^3=\underset{\text{real}}{\underbrace{({\cos }^3\theta -3\cos \theta {\sin }^2\theta )}}+i\underset{\text{imag}}{\underbrace{(3{\cos }^2\theta \sin \theta -{\sin }^3\theta )}}.$ Matching: $\cos 3\theta =4{\cos }^3\theta -3\cos \theta ,\sin 3\theta =3\sin \theta -4{\sin }^3\theta .$ (Using ${\sin }^2\theta =1-{\cos }^2\theta$ to tidy each.)

(No new figure — this is pure algebra riding on Step 4's result.)

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The one-picture summary

Everything above is a single sentence made of pictures: a complex number is an arrow (length $r$, angle $\theta$); powering it $n$ times raises the length to $r^n$ and multiplies the angle to $n\theta$; reversing that (roots) slices the angle into $n$ equal spokes around a circle.

Recall Feynman retelling — the whole walkthrough in plain words

Draw a clock hand of some length pointing at some angle. Multiplying two hands does two simple things: it stretches (lengths multiply) and it spins (angles add) — that's the one fact we proved by multiplying out the brackets and spotting the angle-addition formulas hiding inside. Once you know a single multiply adds one angle, doing it $n$ times just adds the angle $n$ times and stretches $n$ times — that's De Moivre. Multiplying zero times means don't move (that's the $n=0$ case), and a negative power spins backwards (mirror across the flat axis). Finally, to undo the spin — to find where you must start so that $n$ spins land on a target — you cut the target angle into $n$ equal slices, and because a full circle brings you home, there are exactly $n$ starting spots evenly spaced on a ring. Spin adds, stretch multiplies, roots slice the circle. That's the entire story.

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Connections

• Polar form of complex numbers — the $r\mathrm{cis}\theta$ arrow this whole page draws.
• Argand diagram — the graph paper every figure lives on.
• Trigonometric identities — angle-addition formulas power Step 3.
• Euler's formula — the one-line "cheat" version of the same result.
• Roots of unity — Step 7's equally spaced spokes.
• Binomial theorem — Step 8's expansion for multiple angles.