De Moivre's Theorem

De Moivre's Theorem is one of the most important results in complex number theory. It provides a simple formula for raising a complex number to any integer power and has numerous applications, including finding trigonometric identities and computing roots of complex numbers.

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Statement of De Moivre's Theorem

For any integer $n$n (positive, negative, or zero) and any angle $\theta$θ:

$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$(cosθ+isinθ)n=cos(nθ)+isin(nθ)

Equivalently, if $z = r(\cos\theta + i\sin\theta)$z=r(cosθ+isinθ), then:

$z^n = r^n(\cos(n\theta) + i\sin(n\theta))$zn=rn(cos(nθ)+isin(nθ))

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Proof by Induction (for positive integers)

Base case: $n = 1$n=1: $(\cos\theta + i\sin\theta)^1 = \cos\theta + i\sin\theta$(cosθ+isinθ)1=cosθ+isinθ ✓

Inductive step: Assume the result holds for $n = k$n=k:

$(\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta)$(cosθ+isinθ)k=cos(kθ)+isin(kθ)

Then for $n = k + 1$n=k+1:

$(\cos\theta + i\sin\theta)^{k+1} = (\cos(k\theta) + i\sin(k\theta))(\cos\theta + i\sin\theta)$(cosθ+isinθ)k+1=(cos(kθ)+isin(kθ))(cosθ+isinθ)

Using the multiplication rule (addition formulae):

$= \cos(k\theta + \theta) + i\sin(k\theta + \theta) = \cos((k+1)\theta) + i\sin((k+1)\theta)$=cos(kθ+θ)+isin(kθ+θ)=cos((k+1)θ)+isin((k+1)θ)

By induction, the result holds for all positive integers $n$n.

For negative integers, we use $z^{-n} = \frac{1}{z^n}$z−n=zn1​ and the result follows similarly.

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Computing Powers of Complex Numbers

Worked Example 1: Find $(1 + i)^{10}$(1+i)10.

Step 1: Convert to polar form. $|1 + i| = \sqrt{2}$∣1+i∣=2​, $\arg(1 + i) = \frac{\pi}{4}$arg(1+i)=4π​.

$1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$1+i=2​(cos4π​+isin4π​)

Step 2: Apply De Moivre's Theorem.

$(1 + i)^{10} = (\sqrt{2})^{10}\left(\cos\frac{10\pi}{4} + i\sin\frac{10\pi}{4}\right)$(1+i)10=(2​)10(cos410π​+isin410π​)

$= 32\left(\cos\frac{5\pi}{2} + i\sin\frac{5\pi}{2}\right)$=32(cos25π​+isin25π​)

$= 32\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) \quad (\text{since } \frac{5\pi}{2} = 2\pi + \frac{\pi}{2})$=32(cos2π​+isin2π​)(since 25π​=2π+2π​)

$= 32(0 + i) = 32i$=32(0+i)=32i

Worked Example 2: Find $(\sqrt{3} - i)^6$(3​−i)6.

$|\sqrt{3} - i| = 2$∣3​−i∣=2, $\arg = -\frac{\pi}{6}$arg=−6π​ (fourth quadrant).

$(\sqrt{3} - i)^6 = 2^6 \left(\cos\left(-\pi\right) + i\sin\left(-\pi\right)\right) = 64(-1 + 0) = -64$(3​−i)6=26(cos(−π)+isin(−π))=64(−1+0)=−64

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Deriving Trigonometric Identities

De Moivre's Theorem is a powerful tool for deriving multiple-angle formulae.

Method: Expanding $(\cos\theta + i\sin\theta)^n$(cosθ+isinθ)n

By De Moivre's Theorem, the left side equals $\cos(n\theta) + i\sin(n\theta)$cos(nθ)+isin(nθ).

By the binomial theorem, expand the left side and equate real and imaginary parts.

Worked Example 3: Express $\cos(3\theta)$cos(3θ) and $\sin(3\theta)$sin(3θ) in terms of $\cos\theta$cosθ and $\sin\theta$sinθ.

$(\cos\theta + i\sin\theta)^3 = \cos(3\theta) + i\sin(3\theta)$(cosθ+isinθ)3=cos(3θ)+isin(3θ)

Expand the left side using the binomial theorem:

$\cos^3\theta + 3\cos^2\theta(i\sin\theta) + 3\cos\theta(i\sin\theta)^2 + (i\sin\theta)^3$cos3θ+3cos2θ(isinθ)+3cosθ(isinθ)2+(isinθ)3

$= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta$=cos3θ+3icos2θsinθ−3cosθsin2θ−isin3θ

$= (\cos^3\theta - 3\cos\theta\sin^2\theta) + i(3\cos^2\theta\sin\theta - \sin^3\theta)$=(cos3θ−3cosθsin2θ)+i(3cos2θsinθ−sin3θ)

Equating real parts:

$\cos(3\theta) = \cos^3\theta - 3\cos\theta\sin^2\theta = 4\cos^3\theta - 3\cos\theta$cos(3θ)=cos3θ−3cosθsin2θ=4cos3θ−3cosθ

Equating imaginary parts: