CP 2.8 - Sums of series with de Moivre | Edexcel A-Level Further Mathematics

Sum trigonometric series by treating their cosine and sine terms as the real and imaginary parts of one complex geometric progression. You will derive finite-sum formulae, prove a useful half-circle identity and handle geometric weights as well as unit-length terms.

Summing a complex geometric progression

Can you find

1+\cos\frac{\pi}{20}+\cos\frac{2\pi}{20}+\cdots+\cos\frac{19\pi}{20}

without calculating nineteen cosines? Its exact value is $1$ . There is also a compact exact expression for the matching sine sum. A single complex series will produce both answers.

You already know

From A-level Mathematics, a geometric progression has a constant ratio. De Moivre gives $q^k=\cos k\theta+\mathrm{i}\sin k\theta$ when $q=\cos\theta+\mathrm{i}\sin\theta$ . Thus the real and imaginary parts of $1+q+\cdots+q^{n-1}$ are the corresponding cosine and sine sums.

For example, let $q=\cos\frac{\pi}{3}+\mathrm{i}\sin\frac{\pi}{3}$ . The terms $1,q,q^2,\ldots$ are unit arrows whose directions turn by $\frac{\pi}{3}$ each time. Adding them means placing the arrows head to tail. Six close the chain, while four finish on the imaginary axis.

[DIAGRAM: asset_name: A complex geometric sum as a chain of arrows; asset_slug: 9fm0-cp11-geometric-chain; caption: Six unit terms close the chain and four have sum i root 3; description: Six successive unit arrows, directed at zero, pi over 3, two pi over 3, pi, four pi over 3 and five pi over 3, form a closed hexagonal chain. The first four are blue and the last two sage. The endpoint after four terms is marked i root 3. A side panel gives q equals cosine pi over 3 plus i sine pi over 3 and explains that the six-term sum is zero.]
Six unit terms close the chain and four have sum i root 3

The familiar finite geometric-series formula works for a complex ratio because its proof uses only multiplication and subtraction. Write $T_n=1+q+\cdots+q^{n-1}$ ; then $qT_n=q+q^2+\cdots+q^n$ . Subtraction gives $(1-q)T_n=1-q^n$ .

Finite geometric series

For a positive integer $n$ and complex $q\ne1$ ,

T_n=1+q+q^2+\cdots+q^{n-1}=\frac{1-q^n}{1-q}

For $q=1$ , add the $n$ terms directly to obtain $T_n=n$ . The finite-sum formula has no restriction $|q|<1$ .

In the formula booklet.

Count the terms carefully: the last exponent is $n-1$ , but the numerator contains $q^n$ . The first example uses four terms, so it needs the fourth power in the numerator.

Worked example 1: four unit terms

Question. For $q=\cos\frac{\pi}{3}+\mathrm{i}\sin\frac{\pi}{3}$ , find $1+q+q^2+q^3$ in Cartesian form.

Plan. Apply the finite GP formula, evaluate $q^4$ using de Moivre and rationalise the denominator.

\begin{aligned} q&=\frac12+\frac{\mathrm{i}\sqrt{3}}2 && \text{exact values} \\ q^4&=\cos\frac{4\pi}{3}+\mathrm{i}\sin\frac{4\pi}{3} =-\frac12-\frac{\mathrm{i}\sqrt{3}}2 && \text{de Moivre} \\ T_4&=\frac{1-q^4}{1-q} =\frac{\frac32+\frac{\mathrm{i}\sqrt{3}}2}{\frac12-\frac{\mathrm{i}\sqrt{3}}2} && \text{four-term GP} \\ &=\frac{(\frac32+\frac{\mathrm{i}\sqrt{3}}2)(\frac12+\frac{\mathrm{i}\sqrt{3}}2)}{\frac14+\frac34} && \text{multiply by conjugate} \\ &=\mathrm{i}\sqrt{3} && \text{real terms cancel} \end{aligned}

Why this step? The complex denominator is turned into the real number $1$ , so the real and imaginary sums can be read directly.

Answer. $1+q+q^2+q^3=\mathrm{i}\sqrt{3}$ .

Check. The real parts add to $1+\frac12-\frac12-1=0$ ; the imaginary parts add to $0+\frac{\sqrt{3}}2+\frac{\sqrt{3}}2+0=\sqrt{3}$ , agreeing with the diagram.

Change only the number of terms from four to five. The new final term changes both components of the sum.

Worked example 2: five unit terms

Question. For the same $q$ , find $1+q+q^2+q^3+q^4$ in Cartesian form.

Plan. Use $T_5=(1-q^5)/(1-q)$ and simplify with the conjugate.

\begin{aligned} q^5&=\cos\frac{5\pi}{3}+\mathrm{i}\sin\frac{5\pi}{3} =\frac12-\frac{\mathrm{i}\sqrt{3}}2 && \text{de Moivre} \\ T_5&=\frac{\frac12+\frac{\mathrm{i}\sqrt{3}}2}{\frac12-\frac{\mathrm{i}\sqrt{3}}2} && \text{five-term GP} \\ &=\frac{(\frac12+\frac{\mathrm{i}\sqrt{3}}2)^2}{\frac14+\frac34} && \text{multiply by conjugate} \\ &=\frac14-\frac34+\frac{\mathrm{i}\sqrt{3}}2 && \text{expand the square} \end{aligned}

Why this step? The conjugate of the denominator happens to equal the numerator, so the numerator becomes a square; it must still be expanded correctly.

Answer. $T_5=-\frac12+\frac{\mathrm{i}\sqrt{3}}2$ .

Check. Adding $q^4=-\frac12-\frac{\mathrm{i}\sqrt{3}}2$ to $T_4=\mathrm{i}\sqrt{3}$ gives exactly this answer.

The denominator condition is essential; the convergence condition for an infinite GP is a different matter.

Watch out

Wrong: “ $|q|=1$ , so the geometric-series formula cannot be used.” The finite formula needs only $q\ne1$ . The condition $|q|<1$ belongs to a sum to infinity, not to a finite sum.

One more term closes the chain in this example.

Extracting cosine and sine sums

The real part of a complex sum is the sum of its real parts, and the imaginary part is the sum of its imaginary parts. Keeping them together lets one geometric progression do two jobs.

For $q=\cos\theta+\mathrm{i}\sin\theta$ , define

C_n=\sum_{k=0}^{n-1}\cos k\theta \qquad S_n=\sum_{k=0}^{n-1}\sin k\theta

Then $C_n+\mathrm{i}S_n=(1-q^n)/(1-q)$ . The first sine term is zero; the first cosine term is $1$ .

To put that quotient into a useful form, write $u(\phi)=\cos\phi+\mathrm{i}\sin\phi$ . For any real $\phi$ , the double-angle identities give

\begin{aligned} -2\mathrm{i}\sin\phi\,u(\phi) &=2\sin^2\phi-2\mathrm{i}\sin\phi\cos\phi \\ &=1-\cos2\phi-\mathrm{i}\sin2\phi \\ &=1-u(2\phi) \end{aligned}

Apply this with $\phi=n\theta/2$ in the numerator and $\phi=\theta/2$ in the denominator. The common factor $-2\mathrm{i}$ cancels, while dividing the two unit complex numbers subtracts their angles:

\frac{1-q^n}{1-q} =\frac{\sin(n\theta/2)}{\sin(\theta/2)} \left(\cos\frac{(n-1)\theta}{2}+\mathrm{i}\sin\frac{(n-1)\theta}{2}\right)

The derivation is still valid when $\sin(n\theta/2)=0$ : we never divided by that numerator factor.

Finite cosine and sine sums

For positive integer $n$ and real $\theta$ with $\sin(\theta/2)\ne0$ ,

C_n=\frac{\sin(n\theta/2)\cos((n-1)\theta/2)}{\sin(\theta/2)}

S_n=\frac{\sin(n\theta/2)\sin((n-1)\theta/2)}{\sin(\theta/2)}

If $\theta$ is an integer multiple of $2\pi$ , use $C_n=n$ and $S_n=0$ directly. You may derive these sums from the finite GP instead of memorising them.

Not in the formula booklet: learn it.

The angles in the factors have different jobs: $n\theta/2$ comes from the numerator, whereas $(n-1)\theta/2$ is the middle direction of the terms being added.

Worked example 3: five equally spaced angles

Question. Find the exact values of $\sum_{k=0}^{4}\cos\frac{k\pi}{6}$ and $\sum_{k=0}^{4}\sin\frac{k\pi}{6}$ .

Plan. Use $n=5$ , $\theta=\frac{\pi}{6}$ in the two finite-sum formulae.

\begin{aligned} \frac{\sin(5\pi/12)}{\sin(\pi/12)} &=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}} && \text{exact angle values} \\ &=2+\sqrt{3} && \text{rationalise} \end{aligned}

Therefore

\begin{aligned} C_5&=(2+\sqrt{3})\cos\frac{\pi}{3} =1+\frac{\sqrt{3}}2 && \text{real part} \\ S_5&=(2+\sqrt{3})\sin\frac{\pi}{3} =\frac32+\sqrt{3} && \text{imaginary part} \end{aligned}

Why this step? Five terms run from angle $0$ to $4\theta$ , so their middle direction is $2\theta=\frac{\pi}{3}$ , not $\frac{5\pi}{12}$ .

Answer. The cosine sum is $1+\frac{\sqrt{3}}2$ , and the sine sum is $\frac32+\sqrt{3}$ .

Check. Directly, the cosine terms are $1,\frac{\sqrt{3}}2,\frac12,0,-\frac12$ , and the sine terms are $0,\frac12,\frac{\sqrt{3}}2,1,\frac{\sqrt{3}}2$ .

Change only the number of terms from five to six, keeping the angular step $\frac{\pi}{6}$ .

Worked example 4: six equally spaced angles

Question. Find the exact values of $\sum_{k=0}^{5}\cos\frac{k\pi}{6}$ and $\sum_{k=0}^{5}\sin\frac{k\pi}{6}$ .

Plan. Use $n=6$ and the same $\theta$ , then simplify the real and imaginary parts.

Since $1/\sin\frac{\pi}{12}=\sqrt{6}+\sqrt{2}$ ,

\begin{aligned} C_6&=\frac{\sin(\pi/2)\cos(5\pi/12)}{\sin(\pi/12)} && \text{cosine-sum formula} \\ &=(\sqrt{6}+\sqrt{2})\frac{\sqrt{6}-\sqrt{2}}4=1 && \text{difference of squares} \\ S_6&=\frac{\sin(\pi/2)\sin(5\pi/12)}{\sin(\pi/12)} && \text{sine-sum formula} \\ &=\frac{(\sqrt{6}+\sqrt{2})^2}{4}=2+\sqrt{3} && \text{expand the square} \end{aligned}

Why this step? Increasing the number of terms changes both $n\theta/2$ and $(n-1)\theta/2$ ; retaining the old middle direction would mix two different series.

Answer. The cosine sum is $1$ , and the sine sum is $2+\sqrt{3}$ .

Check. The added term has cosine $-\frac{\sqrt{3}}2$ and sine $\frac12$ , taking the previous two answers to these values.

If a series begins at $k=1$ , remove the initial cosine term $1$ from the matching $k=0$ sum. Removing the initial sine term changes nothing.

Watch out

Wrong: $\sum_{k=0}^{n-1}\cos k\theta=\operatorname{Re}(q+q^2+\cdots+q^n)$ . The indices have shifted. Right: use $1+q+\cdots+q^{n-1}$ ; the first term and the number of terms must match the question.

One more term changes the cosine sum but leaves the sine sum unchanged.

The half-circle series identity

A useful special case takes $q=\cos\frac{\pi}{n}+\mathrm{i}\sin\frac{\pi}{n}$ . Raising it to the power $n$ gives $-1$ , so the numerator of the finite GP becomes the real number $2$ .

The following identity is not printed in the formula booklet. Deriving it from the GP also proves the cosine and sine sums together.

Worked example 5: the half-circle identity

Question. Show that, for every positive integer $n$ ,

1+q+q^2+\cdots+q^{n-1}=1+\mathrm{i}\cot\frac{\pi}{2n}

where $q=\cos\frac{\pi}{n}+\mathrm{i}\sin\frac{\pi}{n}$ .

Plan. Use $q^n=-1$ , rationalise $2/(1-q)$ and apply the half-angle identities.

Put $\alpha=\frac{\pi}{n}$ . Then

\begin{aligned} q^n&=\cos\pi+\mathrm{i}\sin\pi=-1 && \text{de Moivre} \\ T_n&=\frac{2}{1-\cos\alpha-\mathrm{i}\sin\alpha} && \text{finite GP, }q\ne1 \\ &=\frac{2(1-\cos\alpha+\mathrm{i}\sin\alpha)}{(1-\cos\alpha)^2+\sin^2\alpha} && \text{multiply by conjugate} \\ &=\frac{1-\cos\alpha+\mathrm{i}\sin\alpha}{1-\cos\alpha} && \text{denominator }2(1-\cos\alpha) \\ &=1+\mathrm{i}\frac{\sin\alpha}{1-\cos\alpha} && \text{split the components} \\ &=1+\mathrm{i}\frac{2\sin(\alpha/2)\cos(\alpha/2)}{2\sin^2(\alpha/2)} && \text{half-angle identities} \\ &=1+\mathrm{i}\cot\frac{\alpha}{2} && \text{cancel a non-zero sine} \end{aligned}

Why this step? $0<\alpha\le\pi$ , so $\sin(\alpha/2)\ne0$ . The cancellation and the GP denominator are both justified, including when $n=1$ .

Answer. $T_n=1+\mathrm{i}\cot\frac{\pi}{2n}$ , as required.

Check. At $n=2$ , $q=\mathrm{i}$ and the sum is $1+\mathrm{i}=1+\mathrm{i}\cot\frac{\pi}{4}$ . At $n=1$ , the sum is $1$ and $\cot\frac{\pi}{2}=0$ .

Change only the angular step, from $\frac{\pi}{n}$ to $\frac{3\pi}{n}$ . Raising the ratio to the power $n$ still gives $-1$ , so nearly all the algebra survives.

Worked example 6: three times the angular step

Question. For positive integer $n$ , let $w=\cos\frac{3\pi}{n}+\mathrm{i}\sin\frac{3\pi}{n}$ . Find $1+w+\cdots+w^{n-1}$ in terms of a cotangent.

Plan. Use the odd half-turn in $w^n$ , then rationalise exactly as before.

Write $\beta=\frac{3\pi}{n}$ . Since $w^n=\cos3\pi+\mathrm{i}\sin3\pi=-1$ ,

\begin{aligned} 1+w+\cdots+w^{n-1} &=\frac{2}{1-\cos\beta-\mathrm{i}\sin\beta} && \text{finite GP} \\ &=1+\mathrm{i}\frac{\sin\beta}{1-\cos\beta} && \text{rationalise} \\ &=1+\mathrm{i}\cot\frac{\beta}{2} && \text{half-angle identities} \end{aligned}

Why this step? A ratio equal to $1$ would have $w^n=1$ , contradicting $w^n=-1$ . Thus the denominator cannot vanish, even when $\beta$ is outside the principal-argument range.

Answer. $\displaystyle 1+w+\cdots+w^{n-1}=1+\mathrm{i}\cot\frac{3\pi}{2n}$ .

Check. At $n=3$ , $w=-1$ , so the sum is $1-1+1=1$ ; the formula gives $1+\mathrm{i}\cot\frac{\pi}{2}=1$ .

The opening puzzle is now solved. Taking real and imaginary parts of Worked example 5 gives

\sum_{k=0}^{n-1}\cos\frac{k\pi}{n}=1 \qquad \sum_{k=0}^{n-1}\sin\frac{k\pi}{n}=\cot\frac{\pi}{2n}

These specialised sums are not printed in the formula booklet. At $n=20$ , the cosine sum is exactly $1$ , and the sine sum is $\cot\frac{\pi}{40}$ . The cosine terms other than the initial $1$ cancel in pairs, because $\cos(\pi-\phi)=-\cos\phi$ ; the complex method discovers the less obvious sine sum at the same time.

Watch out

Wrong: $q^n=\cos\pi+\mathrm{i}\sin\pi=1$ . Right: it is $-1$ . A half-turn gives the numerator $1-(-1)=2$ ; a full turn gives numerator $0$ .

An odd multiple of the same angular step leads to the same pattern, with a changed cotangent argument.

Geometrically weighted trigonometric sums

A geometric weight changes the modulus of the complex ratio. For a real $r>0$ , put $q=r(\cos\theta+\mathrm{i}\sin\theta)$ . De Moivre gives $q^k=r^k(\cos k\theta+\mathrm{i}\sin k\theta)$ , so the same finite GP method sums weighted cosine and sine terms.

Weighted finite sums

For positive integer $n$ , real $r>0$ , real $\theta$ and $q=r(\cos\theta+\mathrm{i}\sin\theta)\ne1$ ,

\sum_{k=0}^{n-1}r^k\cos k\theta +\mathrm{i}\sum_{k=0}^{n-1}r^k\sin k\theta =\frac{1-q^n}{1-q}

This application of the geometric-sum formula is not printed separately.

Not in the formula booklet: learn it.

Do not use $q+q^{-1}=2\cos\theta$ here unless $r=1$ . The GP formula itself works with either unit or non-unit modulus.

Worked example 7: four weighted terms

Question. Find

A=\sum_{k=0}^{3}\frac{\cos(k\pi/2)}{2^k} \qquad B=\sum_{k=0}^{3}\frac{\sin(k\pi/2)}{2^k}

Plan. Combine the sums using $q=\frac12(\cos\frac{\pi}{2}+\mathrm{i}\sin\frac{\pi}{2})=\frac{\mathrm{i}}2$ .

\begin{aligned} A+\mathrm{i}B &=\frac{1-(\mathrm{i}/2)^4}{1-\mathrm{i}/2} && \text{four-term GP} \\ &=\frac{15/16}{1-\mathrm{i}/2} && (\mathrm{i}/2)^4=1/16 \\ &=\frac{15}{16}\frac{1+\mathrm{i}/2}{1+1/4} && \text{multiply by conjugate} \\ &=\frac34+\frac{3\mathrm{i}}8 && \text{simplify} \end{aligned}

Why this step? The factor $\frac12$ belongs in the complex ratio, so it generates exactly the weights $1,\frac12,\frac14,\frac18$ .

Answer. $A=\frac34$ , $B=\frac38$ .

Check. Directly, $A=1-\frac14=\frac34$ , and $B=\frac12-\frac18=\frac38$ .

Change only the number of terms from four to five. The new term has a real component and no imaginary component.

Worked example 8: five weighted terms

Question. Find $\displaystyle A=\sum_{k=0}^{4}\frac{\cos(k\pi/2)}{2^k}$ and $\displaystyle B=\sum_{k=0}^{4}\frac{\sin(k\pi/2)}{2^k}$ .

Plan. Use $q=\frac{\mathrm{i}}2$ again, with five terms in the finite GP.

\begin{aligned} A+\mathrm{i}B &=\frac{1-(\mathrm{i}/2)^5}{1-\mathrm{i}/2} && \text{five-term GP} \\ &=\frac{1-\mathrm{i}/32}{1-\mathrm{i}/2} && (\mathrm{i}/2)^5=\mathrm{i}/32 \\ &=\frac{(1-\mathrm{i}/32)(1+\mathrm{i}/2)}{5/4} && \text{multiply by conjugate} \\ &=\frac{65/64+15\mathrm{i}/32}{5/4} && \mathrm{i}^2=-1 \\ &=\frac{13}{16}+\frac{3\mathrm{i}}8 && \text{divide both components} \end{aligned}

Why this step? The imaginary part of the numerator is $\frac12-\frac1{32}=\frac{15}{32}$ , while the product of the two imaginary terms adds $\frac1{64}$ to the real part.

Answer. $A=\frac{13}{16}$ , $B=\frac38$ .

Check. The added term is $q^4=\frac1{16}$ , so only the real part increases, from $\frac34$ to $\frac{13}{16}$ .

When the terms have weights, keep those weights in every complex power. Using only a unit complex ratio would sum a different series.

Watch out

Wrong: using $q=\mathrm{i}$ for weights $2^{-k}$ . Right: $q=\frac{\mathrm{i}}2$ . Its modulus controls the weights, and its argument controls the trigonometric angles.

One further term now changes the imaginary sum.

Quick Recap

Combine cosine and sine sums into powers of one complex ratio.
Count the terms: $1+q+\cdots+q^{n-1}$ has $n$ terms and numerator $1-q^n$ .
Use de Moivre for $q^n$ , then simplify the quotient before extracting real and imaginary parts.
The half-circle sum is $1+\mathrm{i}\cot\frac{\pi}{2n}$ ; its real part is always $1$ .
A ratio with modulus $r$ produces geometric weights $r^k$ .

The opening twenty-term cosine sum is exactly $1$ , and its sine partner is $\cot\frac{\pi}{40}$ .

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CP 2.8 - Sums of series with de Moivre

Summing a complex geometric progression

Extracting cosine and sine sums

The half-circle series identity

Geometrically weighted trigonometric sums

Quick Recap

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