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Number and Algebra - De Moivre’s theorem (HL)

Grade 11IB_AA

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, the power is given by zn=rn(cos⁡nθ+isin⁡nθ)z^n = r^n(\cos n\theta + i \sin n\theta).

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In exponential form, De Moivre's Theorem is expressed as (reiθ)n=rneinθ(re^{i\theta})^n = r^n e^{in\theta}.

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The theorem extends to finding roots of complex numbers. The nn roots of z=r(cos⁡θ+isin⁡θ)z = r(\cos \theta + i \sin \theta) are given by z1/n=r1/n(cos⁡θ+2kπn+isin⁡θ+2kπn)z^{1/n} = r^{1/n} \left( \cos \frac{\theta + 2k\pi}{n} + i \sin \frac{\theta + 2k\pi}{n} \right) for k=0,1,2,…,n−1k = 0, 1, 2, \dots, n-1.

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Roots of unity are the solutions to zn=1z^n = 1. They lie on a unit circle in the Argand plane and are separated by an angle of 2πn\frac{2\pi}{n}.

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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 real and imaginary parts.

📐Formulae

[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}

z1/n=r1/ncis(θ+2kπn), for k=0,1,…,n−1z^{1/n} = r^{1/n} \text{cis}\left(\frac{\theta + 2k\pi}{n}\right), \text{ for } k = 0, 1, \dots, n-1

∑k=0n−1ωk=0, where ω is an n-th root of unity (for n>1)\sum_{k=0}^{n-1} \omega^k = 0, \text{ where } \omega \text{ is an } n\text{-th root of unity (for } n > 1)

💡Examples

Problem 1:

Calculate (1+i)10(1 + i)^{10} using De Moivre's Theorem.

Solution:

  1. Convert z=1+iz = 1 + i to polar form: r=12+12=2r = \sqrt{1^2 + 1^2} = \sqrt{2}. θ=arctan⁡(11)=π4\theta = \arctan\left(\frac{1}{1}\right) = \frac{\pi}{4}. So, z=2(cos⁡π4+isin⁡π4)z = \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}).
  2. Apply De Moivre's Theorem: z10=(2)10(cos⁡10π4+isin⁡10π4)z^{10} = (\sqrt{2})^{10} (\cos \frac{10\pi}{4} + i \sin \frac{10\pi}{4}).
  3. Simplify: (2)10=25=32(\sqrt{2})^{10} = 2^5 = 32. 10π4=5π2\frac{10\pi}{4} = \frac{5\pi}{2}. cos⁡5π2=cos⁡π2=0\cos \frac{5\pi}{2} = \cos \frac{\pi}{2} = 0. sin⁡5π2=sin⁡π2=1\sin \frac{5\pi}{2} = \sin \frac{\pi}{2} = 1. z10=32(0+i)=32iz^{10} = 32(0 + i) = 32i.

Explanation:

We first transform the complex number from Cartesian form to polar form to apply the power rule directly. Since 5π2\frac{5\pi}{2} is coterminal with π2\frac{\pi}{2}, the result simplifies to a purely imaginary number.

Problem 2:

Find the three cube roots of z=8iz = 8i.

Solution:

  1. Express 8i8i in polar form: r=8r = 8, θ=π2\theta = \frac{\pi}{2}. z=8(cos⁡π2+isin⁡π2)z = 8(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}).
  2. Use the roots formula z1/3=81/3(cos⁡π/2+2kπ3+isin⁡π/2+2kπ3)z^{1/3} = 8^{1/3} \left( \cos \frac{\pi/2 + 2k\pi}{3} + i \sin \frac{\pi/2 + 2k\pi}{3} \right) 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:

Roots are found by adding multiples of 2π2\pi to the argument before dividing by nn. The resulting roots are equally spaced by 2π3\frac{2\pi}{3} radians around a circle of radius 2.