Summing a Series with De Moivre

Two sums with no common ratio

Take $C=1+\cos \theta +\cos 2\theta +\cdots +\cos n\theta$C = 1 + cos θ + cos 2θ + … + cos nθ and $S=\sin \theta +\sin 2\theta +\cdots +\sin n\theta$S = sin θ + sin 2θ + … + sin nθ. The angle goes up by $\theta$θ each term, but neither sum is geometric: to be geometric, each term would have to be the one before it times the same number.

At $\theta =30^{\circ }$θ = 30°, $\cos 60^{\circ }÷\cos 30^{\circ }\approx 0.577$cos 60° ÷ cos 30° ≈ 0.577, but cos 90° ÷ cos 60° = 0. The ratio changes from one term to the next, so the formula for a geometric sum cannot be used on C, or on S, alone.

Pair them into one complex sum

Add C to i times S, term by term. The terms pair off: C + iS $=1+\left(\cos \theta +i\sin \theta \right)+\left(\cos 2\theta +i\sin 2\theta \right)+\cdots +\left(\cos n\theta +i\sin n\theta \right)$= 1 + (cos θ + i sin θ) + (cos 2θ + i sin 2θ) + … + (cos nθ + i sin nθ).

Let $z=\cos \theta +i\sin \theta$z = cos θ + i sin θ. By De Moivre’s theorem, $z^k=\cos k\theta +i\sin k\theta$zᵏ = cos kθ + i sin kθ for every whole number k, so each bracket is a power of z, and C + iS $=1+z+z^2+\cdots +z^n$= 1 + z + z² + … + zⁿ. That sum is geometric: each term is the one before it times z.

The terms 1, z, $z^2$z² and $z^3$z³ at $\theta =30^{\circ }$θ = 30°, laid head to tail. Each has length 1 and turns 30° from the one before. The gold arrow is their sum, C + iS: its real part is $C=1+\cos 30^{\circ }+\cos 60^{\circ }+\cos 90^{\circ }\approx 2.366$C = 1 + cos 30° + cos 60° + cos 90° ≈ 2.366, and its imaginary part is $S\approx 2.366$S ≈ 2.366.

Sum it as a geometric series

A geometric series with first term 1 and ratio z has the sum $1+z+z^2+\cdots +z^n=\frac{1-z^{n+1}}{1-z}$1 + z + z² + … + zⁿ = (1 − zⁿ⁺¹)/(1 − z), as long as $z\ne 1$z ≠ 1. There are n + 1 terms, from $z^0$z⁰ up to $z^n$zⁿ, so the power on top is n + 1.

So C + iS $=\frac{1-z^{n+1}}{1-z}$= (1 − zⁿ⁺¹)/(1 − z), and $z^{n+1}=\cos \left(\left(n+1\right)\theta \right)+i\sin \left(\left(n+1\right)\theta \right)$zⁿ⁺¹ = cos((n + 1)θ) + i sin((n + 1)θ) by De Moivre again.

Split the quotient into its parts

C is the real part of the quotient and S is its imaginary part. To read them off, write the top and the bottom in polar form.

Use the double-angle formulas $1-\cos \theta =2{\sin }^2\left(\frac{\theta }{2}\right)$1 − cos θ = 2sin²(θ/2) and $\sin \theta =2\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)$sin θ = 2sin(θ/2) cos(θ/2). Then $1-z=2{\sin }^2\left(\frac{\theta }{2}\right)-2i\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)=-2i\sin \left(\frac{\theta }{2}\right)\left[\cos \left(\frac{\theta }{2}\right)+i\sin \left(\frac{\theta }{2}\right)\right]$1 − z = 2sin²(θ/2) − 2i sin(θ/2) cos(θ/2) = −2i sin(θ/2) [cos(θ/2) + i sin(θ/2)]. Multiply out the right side to check: $-2i\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)-2i^2{\sin }^2\left(\frac{\theta }{2}\right)=2{\sin }^2\left(\frac{\theta }{2}\right)-2i\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)$−2i sin(θ/2) cos(θ/2) − 2i² sin²(θ/2) = 2sin²(θ/2) − 2i sin(θ/2) cos(θ/2).

The top works the same way with $\left(n+1\right)\theta$(n + 1)θ in place of $\theta$θ: $1-z^{n+1}=-2i\sin \left(\frac{\left(n+1\right)\theta }{2}\right)\left[\cos \left(\frac{\left(n+1\right)\theta }{2}\right)+i\sin \left(\frac{\left(n+1\right)\theta }{2}\right)\right]$1 − zⁿ⁺¹ = −2i sin((n + 1)θ/2) [cos((n + 1)θ/2) + i sin((n + 1)θ/2)].

Divide. The −2i cancels, and dividing the two brackets subtracts their angles: $\frac{\left(n+1\right)\theta }{2}-\frac{\theta }{2}=\frac{n\theta }{2}$(n + 1)θ/2 − θ/2 = nθ/2. So C + iS $=\left[\sin \left(\frac{\left(n+1\right)\theta }{2}\right)/\sin \left(\frac{\theta }{2}\right)\right]\times \left[\cos \left(\frac{n\theta }{2}\right)+i\sin \left(\frac{n\theta }{2}\right)\right]$= [sin((n + 1)θ/2) / sin(θ/2)] × [cos(nθ/2) + i sin(nθ/2)].

The first factor is real, so the parts can be read off: $C=\sin \left(\frac{\left(n+1\right)\theta }{2}\right)\cos \left(\frac{n\theta }{2}\right)/\sin \left(\frac{\theta }{2}\right)$C = sin((n + 1)θ/2) cos(nθ/2) / sin(θ/2) and $S=\sin \left(\frac{\left(n+1\right)\theta }{2}\right)\sin \left(\frac{n\theta }{2}\right)/\sin \left(\frac{\theta }{2}\right)$S = sin((n + 1)θ/2) sin(nθ/2) / sin(θ/2).

Checks

At $\theta =60^{\circ }$θ = 60° with n = 2: $C=1+\cos 60^{\circ }+\cos 120^{\circ }=1+\frac{1}{2}-\frac{1}{2}=1$C = 1 + cos 60° + cos 120° = 1 + 1/2 − 1/2 = 1. The formula gives $\sin 90^{\circ }\cos 60^{\circ }÷\sin 30^{\circ }=1\times \frac{1}{2}÷\frac{1}{2}=1$sin 90° cos 60° ÷ sin 30° = 1 × 1/2 ÷ 1/2 = 1. And $S=\sin 60^{\circ }+\sin 120^{\circ }=\sqrt{3}$S = sin 60° + sin 120° = √3, while the formula gives $\sin 90^{\circ }\sin 60^{\circ }÷\sin 30^{\circ }=\left(\frac{\sqrt{3}}{2}\right)/\left(\frac{1}{2}\right)=\sqrt{3}$sin 90° sin 60° ÷ sin 30° = (√3/2) / (1/2) = √3.

At $\theta =30^{\circ }$θ = 30° with n = 3, the sum drawn above: $C=\sin 60^{\circ }\cos 45^{\circ }÷\sin 15^{\circ }\approx 0.866\times 0.707/0.259\approx 2.366$C = sin 60° cos 45° ÷ sin 15° ≈ 0.866 × 0.707 / 0.259 ≈ 2.366, and term by term $C=1+\frac{\sqrt{3}}{2}+\frac{1}{2}+0=\frac{3+\sqrt{3}}{2}\approx 2.366$C = 1 + √3/2 + 1/2 + 0 = (3 + √3)/2 ≈ 2.366.

At $\theta =90^{\circ }$θ = 90° every term cancels

At $\theta =90^{\circ }$θ = 90°, z = cos 90° + i sin 90° = i. With n = 3 the sum is $1+i+i^2+i^3=1+i-1-i=0$1 + i + i² + i³ = 1 + i − 1 − i = 0, and the formula agrees: $\frac{1-i^4}{1-i}=\frac{1-1}{1-i}=0$(1 − i⁴)/(1 − i) = (1 − 1)/(1 − i) = 0.

The real parts give C = 1 + cos 90° + cos 180° + cos 270° = 1 + 0 − 1 + 0 = 0, and the imaginary parts give S = 0 + 1 + 0 − 1 = 0. In the closed form, $\sin \left(\frac{\left(n+1\right)\theta }{2}\right)=\sin 180^{\circ }=0$sin((n + 1)θ/2) = sin 180° = 0, so both C and S are 0.

At $\theta =90^{\circ }$θ = 90° the terms 1, i, $i^2$i² and $i^3$i³, laid head to tail, go round a square and end where they started, at 0.

When z = 1

The geometric formula needs $z\ne 1$z ≠ 1. At $\theta =0^{\circ }$θ = 0°, or any whole number of turns, z = 1 and 1 − z = 0. Then every term is 1, so C = n + 1 and S = 0, found by counting rather than by the formula.

At $\theta =120^{\circ }$θ = 120° with n = 2, z is the cube root of 1 called $\omega$ω, and the sum is $1+\omega +{\omega }^2=0$1 + ω + ω² = 0. The closed form gives sin 180° = 0 on top, so C = 0 and S = 0, as The Roots of Unity found.

The usual mistakes

Building the series from $z=\cos \theta$z = cos θ alone. Its powers are ${\cos }^2\theta$cos²θ, ${\cos }^3\theta$cos³θ and so on, and ${\cos }^2\theta$cos²θ is not $\cos 2\theta$cos 2θ: at $\theta =30^{\circ }$θ = 30°, ${\cos }^{2}30^{\circ }=0.75$cos²30° = 0.75 but cos 60° = 0.5. Only the whole of $\cos \theta +i\sin \theta$cos θ + i sin θ has $z^k=\cos k\theta +i\sin k\theta$zᵏ = cos kθ + i sin kθ.

Taking C from the imaginary part. C has no i beside it in C + iS, so C is the real part and S is the imaginary part.

Putting $z^n$zⁿ on top. The sum runs from $z^0$z⁰ to $z^n$zⁿ, which is n + 1 terms, so the top is $1-z^{n+1}$1 − zⁿ⁺¹.

Forgetting the leading 1. At $\theta =90^{\circ }$θ = 90°, 1 + cos 90° + cos 180° + cos 270° is 0, not −1: the first term is cos 0° = 1, and it cancels cos 180° = −1.