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De Moivre's theorem and exponential formEdexcel A-Level Further Maths: Revision notes

Section 1

De Moivre's theorem

De Moivre's theorem states that for any integer nn, (cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ.(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta. So if z=r(cos⁡θ+isin⁡θ)z=r(\cos\theta+i\sin\theta) then zn=rn(cos⁡nθ+isin⁡nθ)z^n=r^n(\cos n\theta+i\sin n\theta): the modulus is raised to the power nn and the argument is multiplied by nn. It holds for negative nn too, so 1zn=cos⁡nθ−isin⁡nθ\frac{1}{z^n}=\cos n\theta-i\sin n\theta when ∣z∣=1|z|=1. Example: (cos⁡π9+isin⁡π9)6=cos⁡2π3+isin⁡2π3=−12+32i\left(\cos\frac{\pi}{9}+i\sin\frac{\pi}{9}\right)^6=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}=-\frac12+\frac{\sqrt3}{2}i.

Key termsde Moivre's theorem
Common mistake

Forgetting to raise the modulus to the power nn as well as multiplying the argument by nn.

Section 2

Multiple angle formulae

To express cos⁡pθ\cos p\theta or sin⁡qθ\sin q\theta in powers of sin⁡θ\sin\theta and cos⁡θ\cos\theta: expand (cos⁡θ+isin⁡θ)n(\cos\theta+i\sin\theta)^n using the binomial theorem, then equate real and imaginary parts with cos⁡nθ+isin⁡nθ\cos n\theta+i\sin n\theta. For n=3n=3: (cos⁡θ+isin⁡θ)3=cos⁡3θ−3cos⁡θsin⁡2θ+i(3cos⁡2θsin⁡θ−sin⁡3θ)(\cos\theta+i\sin\theta)^3=\cos^3\theta-3\cos\theta\sin^2\theta+i\left(3\cos^2\theta\sin\theta-\sin^3\theta\right). Therefore cos⁡3θ=cos⁡3θ−3cos⁡θsin⁡2θ=4cos⁡3θ−3cos⁡θ\cos3\theta=\cos^3\theta-3\cos\theta\sin^2\theta=4\cos^3\theta-3\cos\theta and sin⁡3θ=3cos⁡2θsin⁡θ−sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin3\theta=3\cos^2\theta\sin\theta-\sin^3\theta=3\sin\theta-4\sin^3\theta, using sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1. For tan⁡rθ\tan r\theta, divide the sine result by the cosine result, then divide top and bottom by a power of cos⁡θ\cos\theta: tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}.

Key termsbinomial expansionreal and imaginary parts
Common mistake

Mixing signs in the expansion: the powers of ii cycle i,−1,−i,1i,-1,-i,1.

Exam tip

The real part comes from even powers of isin⁡θi\sin\theta and the imaginary part from odd powers.

Section 3

Powers of sin and cos using z+1/zz+1/z and z−1/zz-1/z

Let z=cos⁡θ+isin⁡θz=\cos\theta+i\sin\theta. Then z+1z=2cos⁡θz+\frac1z=2\cos\theta and z−1z=2isin⁡θz-\frac1z=2i\sin\theta. More generally zn+z−n=2cos⁡nθz^n+z^{-n}=2\cos n\theta and zn−z−n=2isin⁡nθz^n-z^{-n}=2i\sin n\theta. To write cos⁡pθ\cos^p\theta or sin⁡qθ\sin^q\theta in multiple angles: expand (z±1z)p\left(z\pm\frac1z\right)^p with the binomial theorem, then pair zkz^k with z−kz^{-k}. Example: 16cos⁡4θ=(z+1z)4=(z4+z−4)+4(z2+z−2)+6=2cos⁡4θ+8cos⁡2θ+616\cos^4\theta=\left(z+\frac1z\right)^4=\left(z^4+z^{-4}\right)+4\left(z^2+z^{-2}\right)+6=2\cos4\theta+8\cos2\theta+6, so cos⁡4θ=18(cos⁡4θ+4cos⁡2θ+3)\cos^4\theta=\frac18(\cos4\theta+4\cos2\theta+3). For an odd power of sine, (2isin⁡θ)3=−8isin⁡3θ\left(2i\sin\theta\right)^3=-8i\sin^3\theta and the result is sin⁡3θ=14(3sin⁡θ−sin⁡3θ)\sin^3\theta=\frac14(3\sin\theta-\sin3\theta). These forms integrate easily.

Key termspairingpower reduction
Common mistake

Forgetting the factor (2i)q(2i)^q when using z−1zz-\frac1z for sin⁡qθ\sin^q\theta. Remember i4=1i^4=1 but i3=−ii^3=-i.

Exam tip

Check your result at θ=0\theta=0: cos⁡40=1\cos^4 0=1 and 18(1+4+3)=1\frac18(1+4+3)=1.

Section 4

Sums of series

Powers of z=cos⁡θ+isin⁡θz=\cos\theta+i\sin\theta form a geometric series. Use ∑k=0n−1zk=1−zn1−z\sum_{k=0}^{n-1}z^k=\frac{1-z^n}{1-z}, then simplify using zn=cos⁡nθ+isin⁡nθz^n=\cos n\theta+i\sin n\theta. Example (the standard result): for z=cos⁡πn+isin⁡πnz=\cos\frac{\pi}{n}+i\sin\frac{\pi}{n}, zn=−1z^n=-1, so 1+z+⋯+zn−1=21−z1+z+\dots+z^{n-1}=\frac{2}{1-z}. With a=π2na=\frac{\pi}{2n}, 1−z=2sin⁡a(sin⁡a−icos⁡a)1-z=2\sin a(\sin a-i\cos a). Multiplying by sin⁡a+icos⁡asin⁡a+icos⁡a\frac{\sin a+i\cos a}{\sin a+i\cos a} gives 1+icot⁡a1+i\cot a. Real and imaginary parts of such a sum give trigonometric sums, such as ∑cos⁡kθ\sum\cos k\theta.

Key termsgeometric series
Exam tip

Use half-angle forms such as 1−cos⁡2a=2sin⁡2a1-\cos2a=2\sin^2a and sin⁡2a=2sin⁡acos⁡a\sin2a=2\sin a\cos a to factorise 1−z1-z.

Section 5

Exponential form

The definition eiθ=cos⁡θ+isin⁡θe^{i\theta}=\cos\theta+i\sin\theta gives the exponential form z=reiθz=re^{i\theta} with r=∣z∣r=|z| and θ=arg⁡z\theta=\arg z. De Moivre's theorem then reads (eiθ)n=einθ(e^{i\theta})^n=e^{in\theta}, and z1z2=r1r2ei(θ1+θ2)z_1z_2=r_1r_2e^{i(\theta_1+\theta_2)}, z1z2=r1r2ei(θ1−θ2)\frac{z_1}{z_2}=\frac{r_1}{r_2}e^{i(\theta_1-\theta_2)}. Also eiπ=−1e^{i\pi}=-1. Adding and subtracting eiθe^{i\theta} and e−iθe^{-i\theta} gives cos⁡θ=12(eiθ+e−iθ),sin⁡θ=12i(eiθ−e−iθ).\cos\theta=\tfrac12\left(e^{i\theta}+e^{-i\theta}\right),\qquad \sin\theta=\tfrac{1}{2i}\left(e^{i\theta}-e^{-i\theta}\right). Example: z=2eiπ/6z=2e^{i\pi/6} gives z4=16e2πi/3=−8+83 iz^4=16e^{2\pi i/3}=-8+8\sqrt3\,i.

Key termsexponential formEuler's formula
Common mistake

Writing eiθe^{i\theta} with θ\theta in degrees. Use radians.

Exam tip

Exponential form makes products, quotients and powers one line of working.

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Exam questions on De Moivre's theorem and exponential form

  1. Let z=cos⁡π9+isin⁡π9z=\cos\frac{\pi}{9}+i\sin\frac{\pi}{9}.
    Find the exact value of z3+1z3z^3+\frac{1}{z^3}.2 marks
  2. The complex number z=2eiπ/6z=2e^{i\pi/6} is given.
    Find z4z^4 in the form x+iyx+iy, giving exact values.2 marks
  3. Let z=cos⁡θ+isin⁡θz=\cos\theta+i\sin\theta, where θ\theta is real.
    Use de Moivre's theorem to show that sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin3\theta=3\sin\theta-4\sin^3\theta.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).