De Moivre’s Theorem (AQA A-Level Further Maths): Revision Notes

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De Moivre's Theorem

Introduction to de Moivre's theorem

When working with complex numbers in modulus-argument form (also called polar form), you often need to calculate integer powers. De Moivre's theorem provides an elegant way to do this without tedious multiplication.

Consider a complex number written as z=r(cos⁡θ+isin⁡θ)z = r(\cos\theta + i\sin\theta), where rr is the modulus (a positive real number).

infoNote

To understand how powers work, we can use Euler's formula: eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos\theta + i\sin\theta.

This allows us to write:

  • zn=[r(cos⁡θ+isin⁡θ)]nz^n = [r(\cos\theta + i\sin\theta)]^n
  • zn=rn(eiθ)nz^n = r^n(e^{i\theta})^n

Using index laws: rn(eiθ)n=rneinθr^n(e^{i\theta})^n = r^n e^{in\theta}

Converting back to trigonometric form using Euler's formula again: rneinθ=rn(cos⁡(nθ)+isin⁡(nθ))r^n e^{in\theta} = r^n(\cos(n\theta) + i\sin(n\theta))

This gives us de Moivre's theorem.

De Moivre's theorem

Definition: For any complex number in polar form and any integer nn:

[r(cos⁡θ+isin⁡θ)]n=rn(cos⁡(nθ)+isin⁡(nθ))[r(\cos\theta + i\sin\theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta))

chatImportant

Key points:

  • This theorem applies to all integers nn (positive, negative, or zero)
  • Both the modulus and the argument are affected: the modulus is raised to power nn, and the argument is multiplied by nn
  • The complex number must be in modulus-argument form before applying the theorem
  • This result can be proved rigorously using proof by induction

Using de Moivre's theorem to simplify powers

De Moivre's theorem makes calculating powers of complex numbers straightforward. The key is to remember that you need to raise the modulus to the power and multiply the argument by the power.

Converting to Cartesian form (a + bi)

After applying de Moivre's theorem, you often need to convert the result to Cartesian form a+bia + bi. This requires evaluating the trigonometric functions.

chatImportant

Exam Tip: Always check that your complex number is in modulus-argument form before applying de Moivre's theorem. If it's not, convert it first by finding the modulus and argument.

Worked example 1: Powers of complex numbers

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Worked Example: Powers of Complex Numbers

Question (a): Write (cos⁡π3+isin⁡π3)4\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)^4 in the form a+bia + bi.

Solution:

Using de Moivre's theorem with r=1r = 1, θ=π3\theta = \frac{\pi}{3}, and n=4n = 4:

(cos⁡π3+isin⁡π3)4=cos⁡(4×π3)+isin⁡(4×π3)\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)^4 = \cos\left(4 \times \frac{\pi}{3}\right) + i\sin\left(4 \times \frac{\pi}{3}\right)

=cos⁡4π3+isin⁡4π3= \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3}

=−12+i(−32)= -\frac{1}{2} + i\left(-\frac{\sqrt{3}}{2}\right)

=−12−32i= -\frac{1}{2} - \frac{\sqrt{3}}{2}i

Question (b): Write (cos⁡π4−isin⁡π4)6\left(\cos\frac{\pi}{4} - i\sin\frac{\pi}{4}\right)^6 in the form a+bia + bi.

Solution:

First, recognise that cos⁡π4−isin⁡π4=cos⁡(−π4)+isin⁡(−π4)\cos\frac{\pi}{4} - i\sin\frac{\pi}{4} = \cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)

This uses the identities: cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta and sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta

Now apply de Moivre's theorem:

(cos⁡(−π4)+isin⁡(−π4))6=cos⁡(−3π2)+isin⁡(−3π2)\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right)^6 = \cos\left(-\frac{3\pi}{2}\right) + i\sin\left(-\frac{3\pi}{2}\right)

=cos⁡3π2+i(−sin⁡3π2)=0+i(1)=i= \cos\frac{3\pi}{2} + i\left(-\sin\frac{3\pi}{2}\right) = 0 + i(1) = i

chatImportant

Exam Tip: When you see a negative sign in front of the sine term, rewrite it using the negative angle identity before applying de Moivre's theorem.

Worked example 2: Converting from Cartesian form first

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Worked Example: Converting from Cartesian Form

Question: Given the complex number z=−3+iz = -\sqrt{3} + i, use de Moivre's theorem to find z−2z^{-2} in the form a+bia + bi.

Solution:

First, convert to modulus-argument form.

Step 1: Calculate the modulus: ∣z∣=(−3)2+12=3+1=2|z| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2

Step 2: Calculate the argument: tan⁡−1(1−3)=π6\tan^{-1}\left(\frac{1}{-\sqrt{3}}\right) = \frac{\pi}{6}

Since zz is in the second quadrant (negative real part, positive imaginary part): arg⁡(z)=π−π6=5π6\arg(z) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}

Step 3: Write in modulus-argument form: z=2(cos⁡5π6+isin⁡5π6)z = 2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)

Step 4: Apply de Moivre's theorem with n=−2n = -2: z−2=2−2(cos⁡(−2×5π6)+isin⁡(−2×5π6))z^{-2} = 2^{-2}\left(\cos\left(-2 \times \frac{5\pi}{6}\right) + i\sin\left(-2 \times \frac{5\pi}{6}\right)\right)

=14(cos⁡(−5π3)+isin⁡(−5π3))= \frac{1}{4}\left(\cos\left(-\frac{5\pi}{3}\right) + i\sin\left(-\frac{5\pi}{3}\right)\right)

Step 5: Simplify using the fact that −2×5π6=−5π3=−5π3+2π=π3-2 \times \frac{5\pi}{6} = -\frac{5\pi}{3} = -\frac{5\pi}{3} + 2\pi = \frac{\pi}{3} (since angles are equivalent modulo 2π2\pi):

=14(cos⁡π3+isin⁡π3)= \frac{1}{4}\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)

=14(12+i32)= \frac{1}{4}\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)

=18+38i= \frac{1}{8} + \frac{\sqrt{3}}{8}i

chatImportant

Key Reminder: When raising to a negative power, remember to raise the modulus to that power as well, not just multiply the argument.

Proving trigonometric identities using de Moivre's theorem

De Moivre's theorem combined with binomial expansion provides a powerful method for proving trigonometric identities. There are two main types of problems.

Strategy 1: Writing powers of cos θ or sin θ as a series

To write a power of cos⁡θ\cos\theta or sin⁡θ\sin\theta as a series involving cos⁡(nθ)\cos(n\theta) or sin⁡(nθ)\sin(n\theta):

Step 1: Use the exponential forms:

  • cos⁡θ=eiθ+e−iθ2\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}
  • sin⁡θ=eiθ−e−iθ2i\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}

Or equivalently: 2cos⁡θ=eiθ+e−iθ2\cos\theta = e^{i\theta} + e^{-i\theta} and 2isin⁡θ=eiθ−e−iθ2i\sin\theta = e^{i\theta} - e^{-i\theta}

Step 2: Write out the binomial expansion of the power.

Step 3: Use rules of indices to simplify.

Step 4: Group terms together to write the expression in terms of trigonometric functions.

Worked example 3: Proving an identity for cos⁴θ

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Worked Example: Proving an Identity for cos⁴θ

Question: Prove that 8cos⁡4θ≡cos⁡4θ+4cos⁡2θ+38\cos^4\theta \equiv \cos 4\theta + 4\cos 2\theta + 3.

Solution:

Step 1: Start with 2cos⁡θ=eiθ+e−iθ2\cos\theta = e^{i\theta} + e^{-i\theta}, so we want 2cos⁡θ2\cos\theta to start with (to avoid fractions):

16cos⁡4θ≡(2cos⁡θ)4≡(eiθ+e−iθ)416\cos^4\theta \equiv (2\cos\theta)^4 \equiv (e^{i\theta} + e^{-i\theta})^4

Step 2: Write the binomial expansion:

≡(eiθ)4+4(eiθ)3(e−iθ)+6(eiθ)2(e−iθ)2+4(eiθ)(e−iθ)3+(e−iθ)4\equiv (e^{i\theta})^4 + 4(e^{i\theta})^3(e^{-i\theta}) + 6(e^{i\theta})^2(e^{-i\theta})^2 + 4(e^{i\theta})(e^{-i\theta})^3 + (e^{-i\theta})^4

Step 3: Simplify using index rules:

≡e4iθ+4e2iθ+6+4e−2iθ+e−4iθ\equiv e^{4i\theta} + 4e^{2i\theta} + 6 + 4e^{-2i\theta} + e^{-4i\theta}

≡(e4iθ+e−4iθ)+4(e2iθ+e−2iθ)+6\equiv (e^{4i\theta} + e^{-4i\theta}) + 4(e^{2i\theta} + e^{-2i\theta}) + 6

Step 4: Group terms and convert back to trigonometric form:

Recall that e4iθ+e−4iθ=2cos⁡4θe^{4i\theta} + e^{-4i\theta} = 2\cos 4\theta and e2iθ+e−2iθ=2cos⁡2θe^{2i\theta} + e^{-2i\theta} = 2\cos 2\theta

≡2cos⁡4θ+8cos⁡2θ+6\equiv 2\cos 4\theta + 8\cos 2\theta + 6

Therefore:

16cos⁡4θ=2cos⁡4θ+8cos⁡2θ+616\cos^4\theta = 2\cos 4\theta + 8\cos 2\theta + 6

Dividing both sides by 2:

8cos⁡4θ=cos⁡4θ+4cos⁡2θ+38\cos^4\theta = \cos 4\theta + 4\cos 2\theta + 3

Exam Tip: When working with 16cos⁡4θ16\cos^4\theta, remember this is (2cos⁡θ)4(2\cos\theta)^4. The factor of 2 comes from the exponential form and helps avoid dealing with fractions throughout the calculation.

Strategy 2: Writing cos(nθ) or sin(nθ) in terms of powers

To write cos⁡(nθ)\cos(n\theta) or sin⁡(nθ)\sin(n\theta) in terms of powers of cos⁡θ\cos\theta or sin⁡θ\sin\theta:

Step 1: Use de Moivre's theorem: (cos⁡θ+isin⁡θ)n=cos⁡(nθ)+isin⁡(nθ)(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta).

Step 2: Write out the binomial expansion of the left-hand side.

Step 3: Simplify powers of ii.

Step 4: Equate coefficients of real or imaginary parts.

Step 5: Use cos⁡2θ+sin⁡2θ≡1\cos^2\theta + \sin^2\theta \equiv 1 to write the expression as powers of either cos⁡θ\cos\theta or sin⁡θ\sin\theta.

Worked example 4: Expressing sin 5x in terms of powers of sin x

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Worked Example: Expressing sin 5x in Terms of Powers of sin x

Question: Express sin⁡5x\sin 5x in the form Asin⁡x+Bsin⁡3x+Csin⁡5xA\sin x + B\sin^3 x + C\sin^5 x where AA, BB, and CC are constants to be found.

Solution:

Step 1: Use de Moivre's theorem:

cos⁡5x+isin⁡5x≡(cos⁡x+isin⁡x)5\cos 5x + i\sin 5x \equiv (\cos x + i\sin x)^5

Step 2: Write the binomial expansion:

≡cos⁡5x+5cos⁡4x(isin⁡x)+10cos⁡3x(isin⁡x)2+10cos⁡2x(isin⁡x)3+5cos⁡x(isin⁡x)4+(isin⁡x)5\equiv \cos^5 x + 5\cos^4 x(i\sin x) + 10\cos^3 x(i\sin x)^2 + 10\cos^2 x(i\sin x)^3 + 5\cos x(i\sin x)^4 + (i\sin x)^5

Step 3: Simplify powers of ii (remembering i2=−1i^2 = -1, i3=−ii^3 = -i, i4=1i^4 = 1, i5=ii^5 = i):

≡cos⁡5x+5icos⁡4xsin⁡x−10cos⁡3xsin⁡2x−10icos⁡2xsin⁡3x+5cos⁡xsin⁡4x+isin⁡5x\equiv \cos^5 x + 5i\cos^4 x\sin x - 10\cos^3 x\sin^2 x - 10i\cos^2 x\sin^3 x + 5\cos x\sin^4 x + i\sin^5 x

Step 4: Consider only the imaginary parts (since we want sin⁡5x\sin 5x):

Im:sin⁡5x=5cos⁡4xsin⁡x−10cos⁡2xsin⁡3x+sin⁡5x\text{Im}: \sin 5x = 5\cos^4 x\sin x - 10\cos^2 x\sin^3 x + \sin^5 x

Step 5: Replace cos⁡2x\cos^2 x by 1−sin⁡2x1 - \sin^2 x:

sin⁡5x=5(1−sin⁡2x)2sin⁡x−10(1−sin⁡2x)sin⁡3x+sin⁡5x\sin 5x = 5(1 - \sin^2 x)^2\sin x - 10(1 - \sin^2 x)\sin^3 x + \sin^5 x

Expanding:

=5(1−2sin⁡2x+sin⁡4x)sin⁡x−10(1−sin⁡2x)sin⁡3x+sin⁡5x= 5(1 - 2\sin^2 x + \sin^4 x)\sin x - 10(1 - \sin^2 x)\sin^3 x + \sin^5 x

=5sin⁡x−10sin⁡3x+5sin⁡5x−10sin⁡3x+10sin⁡5x+sin⁡5x= 5\sin x - 10\sin^3 x + 5\sin^5 x - 10\sin^3 x + 10\sin^5 x + \sin^5 x

=5sin⁡x−20sin⁡3x+16sin⁡5x= 5\sin x - 20\sin^3 x + 16\sin^5 x

Therefore: A=5A = 5, B=−20B = -20, C=16C = 16

chatImportant

Exam Tip: You can always equate real and imaginary parts in any equation. This is a key technique when working with complex numbers. Also, remember that cos⁡2x\cos^2 x can be replaced by 1−sin⁡2x1 - \sin^2 x when you need to express everything in terms of sin only.

Finding sums of series involving sin(rθ) or cos(rθ)

De Moivre's theorem can be combined with the geometric series formula to find sums of trigonometric series.

Strategy 3: Finding sums of trigonometric series

To find the sum of a series involving sin⁡(rθ)\sin(r\theta) or cos⁡(rθ)\cos(r\theta):

Step 1: Consider a sum involving cos⁡(rθ)+isin⁡(rθ)\cos(r\theta) + i\sin(r\theta).

Step 2: Use de Moivre's theorem to rewrite as (cos⁡θ+isin⁡θ)r(\cos\theta + i\sin\theta)^r.

Step 3: Use the formula for the sum of a geometric series:

S=a(1−rn)1−rS = \frac{a(1-r^n)}{1-r}

where aa is the first term and rr is the common ratio.

Step 4: Use formulae for sin⁡(A±B)\sin(A \pm B) or cos⁡(A±B)\cos(A \pm B) to simplify.

Step 5: Select only the real or the imaginary parts as required.

Worked example 5: Sum of a sine series

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Worked Example: Sum of a Sine Series

Question: Show that ∑r=1nsin⁡(rθ)=sin⁡((n+1)θ2)sin⁡(nθ2)sin⁡(θ2)\displaystyle\sum_{r=1}^{n}\sin(r\theta) = \frac{\sin\left(\frac{(n+1)\theta}{2}\right)\sin\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}

Solution:

Step 1: First consider the sum:

S=∑r=1ncos⁡(rθ)+isin⁡(rθ)S = \sum_{r=1}^{n}\cos(r\theta) + i\sin(r\theta)

Step 2: Use de Moivre's theorem:

=∑r=1n(cos⁡θ+isin⁡θ)r= \sum_{r=1}^{n}(\cos\theta + i\sin\theta)^r

This is a geometric series with:

  • First term: a=cos⁡θ+isin⁡θa = \cos\theta + i\sin\theta
  • Common ratio: r=cos⁡θ+isin⁡θr = \cos\theta + i\sin\theta
  • Number of terms: nn

=(cos⁡θ+isin⁡θ)+(cos⁡θ+isin⁡θ)2+…+(cos⁡θ+isin⁡θ)n= (\cos\theta + i\sin\theta) + (\cos\theta + i\sin\theta)^2 + \ldots + (\cos\theta + i\sin\theta)^n

Step 3: Apply the geometric series formula. We need to find an expression for 1−rn1 - r^n:

1−(cos⁡θ+isin⁡θ)n=1−(cos⁡(nθ)+isin⁡(nθ))1 - (\cos\theta + i\sin\theta)^n = 1 - (\cos(n\theta) + i\sin(n\theta))

=2sin⁡2(nθ2)−2isin⁡(nθ2)cos⁡(nθ2)= 2\sin^2\left(\frac{n\theta}{2}\right) - 2i\sin\left(\frac{n\theta}{2}\right)\cos\left(\frac{n\theta}{2}\right)

=2sin⁡(nθ2)(sin⁡(nθ2)−icos⁡(nθ2))= 2\sin\left(\frac{n\theta}{2}\right)\left(\sin\left(\frac{n\theta}{2}\right) - i\cos\left(\frac{n\theta}{2}\right)\right)

Step 4: For the denominator 1−r1 - r, using addition formulae:

An expression for 1−r1 - r is: 1−(cos⁡θ+isin⁡θ)=2sin⁡(θ2)(sin⁡(θ2)−icos⁡(θ2))1 - (\cos\theta + i\sin\theta) = 2\sin\left(\frac{\theta}{2}\right)\left(\sin\left(\frac{\theta}{2}\right) - i\cos\left(\frac{\theta}{2}\right)\right)

Step 5: Putting these together:

S=(cos⁡θ+isin⁡θ)(1−rn)1−rS = \frac{(\cos\theta + i\sin\theta)(1 - r^n)}{1 - r}

After substituting and simplifying (setting n=1n=1 to help simplification):

=sin⁡(nθ2)(sin⁡(nθ2)−icos⁡(nθ2))(sin⁡(θ2)+icos⁡(θ2))sin⁡(θ2)(sin⁡(θ2)−icos⁡(θ2))(sin⁡(θ2)+icos⁡(θ2))= \frac{\sin\left(\frac{n\theta}{2}\right)\left(\sin\left(\frac{n\theta}{2}\right) - i\cos\left(\frac{n\theta}{2}\right)\right)\left(\sin\left(\frac{\theta}{2}\right) + i\cos\left(\frac{\theta}{2}\right)\right)}{\sin\left(\frac{\theta}{2}\right)\left(\sin\left(\frac{\theta}{2}\right) - i\cos\left(\frac{\theta}{2}\right)\right)\left(\sin\left(\frac{\theta}{2}\right) + i\cos\left(\frac{\theta}{2}\right)\right)}

The denominator simplifies to sin⁡(θ2)\sin\left(\frac{\theta}{2}\right) because: (sin⁡(θ2)−icos⁡(θ2))(sin⁡(θ2)+icos⁡(θ2))=sin⁡2(θ2)+cos⁡2(θ2)=1\left(\sin\left(\frac{\theta}{2}\right) - i\cos\left(\frac{\theta}{2}\right)\right)\left(\sin\left(\frac{\theta}{2}\right) + i\cos\left(\frac{\theta}{2}\right)\right) = \sin^2\left(\frac{\theta}{2}\right) + \cos^2\left(\frac{\theta}{2}\right) = 1

Step 6: The numerator expands to: sin⁡(nθ2)[cos⁡((n−1)θ2)+isin⁡((n−1)θ2)]\sin\left(\frac{n\theta}{2}\right)\left[\cos\left(\frac{(n-1)\theta}{2}\right) + i\sin\left(\frac{(n-1)\theta}{2}\right)\right]

Using product-to-sum formulae and considering the imaginary part:

∑r=1nsin⁡(rθ)=sin⁡((n+1)θ2)sin⁡(nθ2)sin⁡(θ2)\sum_{r=1}^{n}\sin(r\theta) = \frac{\sin\left(\frac{(n+1)\theta}{2}\right)\sin\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}

infoNote

For ∑cos⁡(rθ)\sum\cos(r\theta), you would need to consider the real part of the sum.

chatImportant

Exam Tip: These series sum problems are lengthy. Work carefully through each step, and don't lose track of whether you need the real or imaginary part at the end. The geometric series formula is your key tool here.

Remember!

bookmarkSummary

Key Points to Remember:

  • De Moivre's theorem: [r(cos⁡θ+isin⁡θ)]n=rn(cos⁡(nθ)+isin⁡(nθ))[r(\cos\theta + i\sin\theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta)) for all integers nn.

  • Always convert to modulus-argument form first before applying de Moivre's theorem. Don't forget to raise the modulus to the power as well as multiplying the argument.

  • For proving identities involving powers of trigonometric functions, use the exponential forms 2cos⁡θ=eiθ+e−iθ2\cos\theta = e^{i\theta} + e^{-i\theta} and 2isin⁡θ=eiθ−e−iθ2i\sin\theta = e^{i\theta} - e^{-i\theta}, then apply binomial expansion.

  • When finding sums of series, consider ∑(cos⁡(rθ)+isin⁡(rθ))\sum(\cos(r\theta) + i\sin(r\theta)), recognise it as a geometric series using de Moivre's theorem, then extract the real or imaginary part as needed.

  • Common exam trap: Forgetting to use the negative angle identities when you have subtraction of sine, such as cos⁡θ−isin⁡θ=cos⁡(−θ)+isin⁡(−θ)\cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta).

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