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Number and Algebra - De Moivre's Theorem and its Applications

Grade 12IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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De Moivre's Theorem states that for any complex number z=r(cos⁡θ+isin⁡θ)z = r(\cos \theta + i \sin \theta) and any integer nn, zn=rn(cos⁡nθ+isin⁡nθ)z^n = r^n(\cos n\theta + i \sin n\theta). Visually, this transformation scales the complex vector by a factor of rnr^n and rotates it by an angle nθn\theta around the origin in the Argand plane.

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The polar form z=r(cos⁡θ+isin⁡θ)z = r(\cos \theta + i \sin \theta) and Euler's form z=reiθz = r e^{i\theta} provide a geometric interpretation of complex numbers where rr (the modulus) is the distance from the origin and θ\theta (the argument) is the angle measured counter-clockwise from the positive real axis. On an Argand diagram, multiplying two complex numbers corresponds to multiplying their moduli and adding their arguments.

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The nn-th roots of a complex number w=R(cos⁡ϕ+isin⁡ϕ)w = R(\cos \phi + i \sin \phi) are found by solving zn=wz^n = w. This results in nn distinct roots given by zk=R1/n[cos⁡(ϕ+2kπn)+isin⁡(ϕ+2kπn)]z_k = R^{1/n} [\cos(\frac{\phi + 2k\pi}{n}) + i \sin(\frac{\phi + 2k\pi}{n})] for k=0,1,…,n−1k = 0, 1, \dots, n-1. Visually, these roots are represented as the vertices of a regular nn-sided polygon inscribed in a circle of radius R1/nR^{1/n} centered at the origin.

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The nn-th roots of unity are the solutions to zn=1z^n = 1. These roots lie on the unit circle (radius 1) and are given by ωk=ei2kπn\omega_k = e^{i\frac{2k\pi}{n}}. A key property is that the sum of all nn-th roots of unity is always zero, which can be visualized as the center of mass of the regular nn-gon vertices being at the origin.

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De Moivre's Theorem can be used to derive trigonometric identities. By expanding (cos⁡θ+isin⁡θ)n(\cos \theta + i \sin \theta)^n using the binomial theorem and equating the real and imaginary parts to cos⁡nθ\cos n\theta and sin⁡nθ\sin n\theta respectively, we can express multiple-angle functions in terms of powers of sin⁡θ\sin \theta and cos⁡θ\cos \theta.

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Euler's identity, eiπ+1=0e^{i\pi} + 1 = 0, is a special case of De Moivre's theorem and Euler's form. It represents a rotation of 180∘180^{\circ} on the unit circle, moving from the point (1,0)(1, 0) to (−1,0)(-1, 0) in the complex plane.

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Complex numbers expressed as z=cos⁡θ+isin⁡θz = \cos \theta + i \sin \theta and z−1=cos⁡θ−isin⁡θz^{-1} = \cos \theta - i \sin \theta allow for the expression of cos⁡θ\cos \theta and sin⁡θ\sin \theta in terms of zz. Specifically, 2cos⁡θ=z+1z2\cos \theta = z + \frac{1}{z} and 2isin⁡θ=z−1z2i\sin \theta = z - \frac{1}{z}. This is a powerful tool for integrating powers of trigonometric functions or simplifying complex trigonometric expressions.

📐Formulae

z=r(cos⁡θ+isin⁡θ)=reiθz = r(\cos \theta + i \sin \theta) = r e^{i\theta}

[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)

(eiθ)n=einθ(e^{i\theta})^n = e^{in\theta}

z^{1/n} = \sqrt[n]{r} \left[ \cos\left(\frac{\theta + 2k\pi}{n}\right) + i \sin\left( rac{\theta + 2k\pi}{n}\right) \right], \text{ for } k = 0, 1, \dots, n-1

cos⁡θ=eiθ+e−iheta2\cos \theta = \frac{e^{i\theta} + e^{-i heta}}{2}

sin⁡θ=eiθ−e−iheta2i\sin \theta = \frac{e^{i\theta} - e^{-i heta}}{2i}

zn+1zn=2cos⁡(nθ)z^n + \frac{1}{z^n} = 2\cos(n\theta)

zn−1zn=2isin⁡(nθ)z^n - \frac{1}{z^n} = 2i\sin(n\theta)

💡Examples

Problem 1:

Calculate (3+i)6(\sqrt{3} + i)^6 and express the result in Cartesian form.

Solution:

  1. Find the modulus rr: r=(3)2+12=3+1=2r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2.
  2. Find the argument θ\theta: tan⁡θ=13⇒θ=π6\tan \theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = \frac{\pi}{6}.
  3. Write in polar form: z=2(cos⁡π6+isin⁡π6)z = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}).
  4. Apply De Moivre's Theorem: z6=26(cos⁡(6⋅π6)+isin⁡(6⋅π6))z^6 = 2^6 (\cos(6 \cdot \frac{\pi}{6}) + i \sin(6 \cdot \frac{\pi}{6})).
  5. Simplify: z6=64(cos⁡π+isin⁡π)z^6 = 64 (\cos \pi + i \sin \pi).
  6. Convert to Cartesian: z6=64(−1+i(0))=−64z^6 = 64(-1 + i(0)) = -64.

Explanation:

We first convert the complex number from Cartesian to polar form to make exponentiation easier. Applying De Moivre's Theorem allows us to raise the modulus to the power and multiply the argument. Finally, we evaluate the trigonometric values to return to Cartesian form.

Problem 2:

Find the three cube roots of 8i8i.

Solution:

  1. Express w=8iw = 8i in polar form: r=8r = 8, θ=π2\theta = \frac{\pi}{2}. So, w=8(cos⁡π2+isin⁡π2)w = 8(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}).
  2. Use the root formula: zk=81/3[cos⁡(π/2+2kπ3)+isin⁡(π/2+2kπ3)]z_k = 8^{1/3} [\cos(\frac{\pi/2 + 2k\pi}{3}) + i \sin(\frac{\pi/2 + 2k\pi}{3})] for k=0,1,2k=0, 1, 2.
  3. For k=0k=0: z0=2(cos⁡π6+isin⁡π6)=2(32+i12)=3+iz_0 = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) = 2(\frac{\sqrt{3}}{2} + i\frac{1}{2}) = \sqrt{3} + i.
  4. For k=1k=1: z1=2(cos⁡5π6+isin⁡5π6)=2(−32+i12)=−3+iz_1 = 2(\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6}) = 2(-\frac{\sqrt{3}}{2} + i\frac{1}{2}) = -\sqrt{3} + i.
  5. For k=2k=2: z2=2(cos⁡9π6+isin⁡9π6)=2(cos⁡3π2+isin⁡3π2)=2(0−i)=−2iz_2 = 2(\cos \frac{9\pi}{6} + i \sin \frac{9\pi}{6}) = 2(\cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2}) = 2(0 - i) = -2i.

Explanation:

To find nn-th roots, we represent the number in polar form and include the periodic factor 2kπ2k\pi. This ensures we find all nn distinct roots by incrementing kk. Geometrically, these three points form an equilateral triangle on an Argand diagram.