De Moivre's theorem — statement, proof, applications: Worked examples

Everything here rests on one fact from the parent note. Let me restate it in plain words so no symbol is unearned.

The scenario matrix

Every question De Moivre can throw at you falls into one of these cells. The worked examples below are tagged with the cell they cover, and together they hit all of them.

#	Cell (case class)	What's tricky about it	Example
C1	Positive integer power, base in Q1	plain application	Ex 1
C2	Positive integer power, base in Q2/Q3	must get the argument's sign right	Ex 2
C3	Negative integer power	reciprocal → negative angle	Ex 3
C4	Degenerate base: purely imaginary / purely real	$θ = \pm π /2$ or $0, π$ ; components vanish	Ex 4
C5	Fractional power = $n$ -th roots (all $n$ pieces)	multivalued, spread $2 π / n$ apart, base in Q4	Ex 5
C6	Roots where $w$ itself is negative real	$ϕ = π$ , roots straddle the axes	Ex 6
C7	Multiple-angle identity via binomial	equate real/imag parts	Ex 7
C8	Limiting / summation ( $\sum cis k θ$ , and $θ \to 0$ )	geometric series + a limit	Ex 8
C9	Word problem (rotation in the plane)	translate physical spin into $cis$	Ex 9
C10	Exam twist: a power that lands exactly on the axis, sign trap	reduce angle mod $2 π$	Ex 10

The worked examples

Figure — De Moivre's theorem — statement, proof, applications

Worked example Ex 2 — Positive power, base in Q2 ·

[C2] Compute $(- 1 + i)^{10}$ .

Forecast: length is $2$ , and $- 1 + i$ points up-left (Q2). The danger is reading the angle as $π /4$ instead of $3 π /4$ .

Modulus. $r = (- 1)^{2} + 1^{2} = 2$ .
Argument — mind the quadrant. Real part $< 0$ , imag part $> 0$ ⇒ Q2. The reference angle is $arctan \frac{1}{1} = \frac{π}{4}$ , but in Q2 the true argument is $π - \frac{π}{4} = \frac{3 π}{4}$ . Why the correction? $arctan$ only ever returns Q1/Q4 answers; the naive formula would place the arrow in the wrong half of the plane (see figure — the orange arrow is genuinely in Q2).
Apply. $(- 1 + i)^{10} = (2)^{10} cis (10 \cdot \frac{3 π}{4}) = 2^{5} cis (\frac{30 π}{4}) = 32 cis (\frac{15 π}{2})$ .
Reduce the angle mod $2 π$ . $\frac{15 π}{2} = 7 π + \frac{π}{2}$ . Subtract $2 π$ three times ( $6 π$ ): $\frac{15 π}{2} - 6 π = \frac{3 π}{2}$ . Why? Angles that differ by full turns give the same arrow; reducing keeps the cosine/sine easy.
Evaluate. $cis \frac{3 π}{2} = cos \frac{3 π}{2} + i sin \frac{3 π}{2} = 0 - i = - i$ . Answer $= 32 (- i) = - 32i$ .

Verify: direct binomial or repeated squaring gives $- 32 i$ (checked below). Purely imaginary and negative — consistent with landing on the negative imaginary axis.

Worked example Ex 4 — Degenerate base: purely imaginary ·

[C4] Compute $i^{7}$ and $(- 2)^{5}$ using De Moivre, to see the boundary cases.

Forecast: $i$ is the arrow of length $1$ pointing straight up ( $θ = π /2$ ). $- 2$ points straight left ( $θ = π$ , length $2$ ). The components vanish, so this tests whether De Moivre still behaves.

$i$ in polar form. Real part $= 0$ , imag $= 1$ ⇒ $θ = \frac{π}{2}$ exactly; $r = 1$ . Why not $arctan$ ? Division by zero real-part — you read the angle straight off the axis instead. This is the degenerate case the formula can't compute but the picture can.
$i^{7} = cis (7 \cdot \frac{π}{2}) = cis (\frac{7 π}{2})$ . Reduce: $\frac{7 π}{2} - 2 π = \frac{3 π}{2}$ . So $i^{7} = cis \frac{3 π}{2} = - i$ . Answer $- i$ .
$- 2$ in polar form. $r = 2$ , $θ = π$ (negative real axis). $(- 2)^{5} = 2^{5} cis (5 π)$ . Reduce: $5 π - 4 π = π$ , so $cis (5 π) = - 1$ . Answer $= 32 (- 1) = - 32$ .

Verify: $i^{7} = i^{4} \cdot i^{3} = 1 \cdot (- i) = - i$ ✔; $(- 2)^{5} = - 32$ ✔ (both checked). De Moivre reproduces the ordinary powers exactly on the axes.

Worked example Ex 5 — Fractional power = all roots, base in Q4 ·

[C5] Find all cube roots of $w = 4 - 4 i$ .

Forecast: three arrows of equal length $∣ w ∣^{1/3}$ , spaced $\frac{2 π}{3}$ apart. Since $w$ is in Q4, the first root starts a little below the real axis.

Polar form of $w$ . $∣ w ∣ = 4^{2} + (- 4)^{2} = 32 = 42$ . Real $> 0$ , imag $< 0$ ⇒ Q4, so $ϕ = - \frac{π}{4}$ . Why negative? Below the axis means clockwise angle; $arctan \frac{- 4}{4} = - \frac{π}{4}$ already lives in Q4, so no correction needed here.
Root modulus. $ρ = (42)^{1/3} = (2^{5/2})^{1/3} = 2^{5/6}$ . Why $r^{1/ n}$ ? We need $ρ^{3} = r$ , so take the real cube root of the length.
Root angles. $ψ_{k} = \frac{- \frac{π}{4} + 2 π k}{3}$ , $k = 0, 1, 2$ :
- $k = 0 : ψ_{0} = - \frac{π}{12}$
- $k = 1 : ψ_{1} = - \frac{π}{12} + \frac{2 π}{3} = \frac{7 π}{12}$
- $k = 2 : ψ_{2} = - \frac{π}{12} + \frac{4 π}{3} = \frac{15 π}{12} = \frac{5 π}{4}$
Write them. $z_{k} = 2^{5/6} cis ψ_{k}$ . Why exactly three? Adding $2 π \cdot 3 = 6 π$ inside would divide back to a repeat — so $k = 0, 1, 2$ exhausts the distinct arrows (see Roots of unity).

Verify: each $z_{k}^{3} = (2^{5/6})^{3} cis (3 ψ_{k}) = 42 cis (- \frac{π}{4} + 2 π k) = 4 - 4 i$ . ✔ Their three tips form an equilateral triangle (figure).

Worked example Ex 7 — Multiple-angle identity via binomial ·

[C7] Express $cos 4 θ$ in terms of $cos θ$ only.

Forecast: even multiple ⇒ result should be a polynomial in $cos θ$ of degree $4$ .

Set up. By De Moivre, $cos 4 θ + i sin 4 θ = (cos θ + i sin θ)^{4}$ . Why? De Moivre packages both $cos 4 θ$ and $sin 4 θ$ into one power we can expand.
Binomial expand (see Binomial theorem) with $c = cos θ, s = sin θ$ : $(c + i s)^{4} = c^{4} + 4 c^{3} (i s) + 6 c^{2} (i s)^{2} + 4 c (i s)^{3} + (i s)^{4} .$
Use $i^{2} = - 1, i^{3} = - i, i^{4} = 1$ : $= c^{4} - 6 c^{2} s^{2} + s^{4} + i (4 c^{3} s - 4 c s^{3}) .$
Equate real parts. $cos 4 θ = c^{4} - 6 c^{2} s^{2} + s^{4}$ . Why? Two complex numbers match iff real parts match. Now replace $s^{2} = 1 - c^{2}$ (a Pythagorean identity): $cos 4 θ = c^{4} - 6 c^{2} (1 - c^{2}) + (1 - c^{2})^{2} = 8 c^{4} - 8 c^{2} + 1.$

Verify: $cos 4 θ = 8 cos^{4} θ - 8 cos^{2} θ + 1$ ; plug $θ = \frac{π}{3}$ : LHS $cos \frac{4 π}{3} = - \frac{1}{2}$ , RHS $= 8 (\frac{1}{16}) - 8 (\frac{1}{4}) + 1 = \frac{1}{2} - 2 + 1 = - \frac{1}{2}$ . ✔

Worked example Ex 8 — Summation + limit ·

[C8] Evaluate $S = k = 0 \sum n - 1 cos k θ$ and check its $θ \to 0$ limit.

Forecast: as $θ \to 0$ every cosine $\to 1$ , so the sum should $\to n$ .

Complexify. Let $C + i S = \sum_{k = 0}^{n - 1} cis (k θ) = \sum_{k = 0}^{n - 1} (cis θ)^{k}$ . Why? $cis (k θ) = (cis θ)^{k}$ by De Moivre, turning the sum into a geometric series with ratio $cis θ$ .
Geometric sum (ratio $\neq = 1$ ): $k = 0 \sum n - 1 q^{k} = \frac{q ^{n} - 1}{q - 1}$ with $q = cis θ$ , giving $\frac{cis ( n θ ) - 1}{cis θ - 1}$ .
Standard closed form (after the half-angle simplification): the real part is $C = \sum_{k = 0}^{n - 1} cos k θ = \frac{s i n \frac{n θ}{2}}{s i n \frac{θ}{2}} cos \frac{( n - 1 ) θ}{2} .$
Limit $θ \to 0$ . Both $sin$ 's $\to$ their arguments: $\frac{s i n ( n θ /2 )}{s i n ( θ /2 )} \to \frac{n θ /2}{θ /2} = n$ , and $cos \to 1$ . So $C \to n$ . Why is this the right sanity check? Setting $θ = 0$ makes each term $cos 0 = 1$ , and there are $n$ terms — the formula must recover that.

Verify: at $θ = \frac{π}{2}, n = 4$ : direct sum $cos 0 + cos \frac{π}{2} + cos π + cos \frac{3 π}{2} = 1 + 0 - 1 + 0 = 0$ ; formula gives $\frac{s i n π}{s i n \frac{π}{4}} cos \frac{3 π}{4} = 0$ . ✔

Worked example Ex 9 — Word problem: rotating a point ·

[C9] A drone sits at the plane point $P = (3, 1)$ relative to its base. Its controller commands it to orbit the base by $9 0^{\circ}$ anticlockwise, three times in a row. Where does it end up?

Forecast: three $9 0^{\circ}$ turns = one $27 0^{\circ}$ turn. The point should swing from Q1 down into Q4.

Encode position as a complex number. $P \leftrightarrow z = 3 + i$ . Why complex? Rotating a point about the origin by angle $α$ is exactly multiplying by $cis α$ — the Argand diagram's spin rule.
One $9 0^{\circ}$ turn = multiply by $cis \frac{π}{2} = i$ . Three turns = multiply by $i^{3}$ . By De Moivre, $(cis \frac{π}{2})^{3} = cis \frac{3 π}{2} = - i$ .
Apply. $z^{'} = (3 + i) \cdot (- i) = - 3 i - i^{2} = 1 - 3 i$ . So the drone lands at $(1, - 3)$ .

Verify: length preserved: $∣3 + i ∣ = 10 = ∣1 - 3 i ∣$ ✔ (a rotation cannot change distance from base). Direction swung $27 0^{\circ}$ : Q1 $\to$ Q4, matching $(1, - 3)$ . (checked below).

Worked example Ex 10 — Exam twist: lands exactly on an axis ·

[C10] Simplify $(\frac{1 + i}{2})^{2026}$ .

Forecast: the base has length exactly $1$ (it's $cis \frac{π}{4}$ ), so no stretching — only spinning. The whole game is reducing $2026 \cdot \frac{π}{4}$ mod $2 π$ . Watch the sign.

Recognise unit modulus. $\frac{1 + i}{2}$ has modulus $\frac{2}{2} = 1$ and argument $\frac{π}{4}$ , i.e. it equals $cis \frac{π}{4}$ . Why check the modulus first? If length $= 1$ , the answer stays on the unit circle — the trap is thinking it blows up.
Apply De Moivre. $(cis \frac{π}{4})^{2026} = cis (\frac{2026 π}{4}) = cis (\frac{1013 π}{2})$ .
Reduce mod $2 π$ . $\frac{1013 π}{2} = 506 π + \frac{π}{2}$ . Since $506 π = 253 \cdot 2 π$ (a whole number of turns), it drops out: angle $\equiv \frac{π}{2}$ . Why does an even multiple of $π$ vanish? $2 π$ is one full turn back home.
Evaluate. $cis \frac{π}{2} = i$ . Answer $= i$ .

Verify: $2026 mod 8 = 2$ (since $cis \frac{π}{4}$ has period $8$ under powering), and $(cis \frac{π}{4})^{2} = cis \frac{π}{2} = i$ . ✔ (checked below).

Recall Self-test across the matrix

Which example proves De Moivre still works when the base sits on an axis? ::: Ex 4 ( $i^{7}$ and $(- 2)^{5}$ ), the degenerate cases. In Ex 2, why is the argument $\frac{3 π}{4}$ not $\frac{π}{4}$ ? ::: The base is in Q2 (real part negative), so add the reference angle to $π$ . How many fourth roots does $- 16$ have and what shape do they make? ::: Four, at the corners of a square of radius $2$ . What does multiplying a point by $cis α$ do physically (Ex 9)? ::: Rotates it about the origin by $α$ , keeping its distance fixed.

Connections

Parent topic — the statement and proof this page exercises.
Polar form of complex numbers — every example starts by converting to $r cis θ$ .
Roots of unity — Ex 5 and Ex 6 are the general- $w$ version.
Euler's formula — the shortcut behind all of this.
Binomial theorem — the engine of Ex 7.
Trigonometric identities — Pythagorean swap used in Ex 7, half-angle in Ex 8.
Argand diagram — the spin-and-stretch picture, central to Ex 9.