De Moivre's Formula

De Moivre's Theorem states that for any complex number xx and integer nn,

(cosx+isinx)n=cos(nx)+isin(nx) ( \cos x + i \sin x )^n = \cos ( nx) + i \sin (nx)

This formula is easily proven by induction on nn, and applying the trigonometric sum and product formulas. We first deal with the non-negative integers. The base case n=0n=0 is clearly true. For the induction step, observe that

(cosx+isinx)k+1=(cosx+isinx)k×(cosx+isinx)=(cos(kx)+isin(kx))(cosx+isinx)=cos(kx)cosxsin(kx)sinx+i[sin(kx)cosx+cos(kx)sinx]=cos[(k+1)x]+isin[(k+1)x] \begin{array} { l l } ( \cos x + i \sin x)^{k+1} & = (\cos x + i \sin x )^k \times ( \cos x + i \sin x ) \\ & = \left( \cos (kx) + i \sin (kx) \right) ( \cos x + i \sin x ) \\ & = \cos (kx) \cos x - \sin(kx) \sin x + i[ \sin (kx) \cos x + \cos(kx) \sin x] \\ & = \cos [(k+1)x] + i \sin [(k+1)x]\\ \end{array}

Note that this version of the proof only holds true for integers nn. There is a more general version, in which nn is allowed to be a complex number. In this case, the left hand side is a multi-valued function, and the right hand side is one of its possible values.

Euler's Formula for complex numbers states that if zz is a complex number with absolute value rz r_z and argument θz \theta_z , then

z=rzeiθz. z = r_z e^{i \theta_z}.

The proof of this is best approached using the (Maclaurin) power series expansion, and is left to the interested reader. With this, we have another proof of De Moivre's theorem that directly follows from the muliplication of complex numbers in polar form.

Worked Examples

1. Evaluate (1+3i)2013 ( 1 + \sqrt{3} i )^{2013} .

First, we find the absolute value and argument of z=1+3i z = 1 + \sqrt{3} i .
Absolute value: rz=12+32=4=2 r_z = \sqrt{ 1^2 + \sqrt{3} ^2 } = \sqrt{4} = 2 .
Argument: θz=arctan31=π3 \theta_z = \arctan \frac{\sqrt{3} } {1} = \frac{\pi}{3}

Applying DeMoivre's Theorem, we obtain

[2 cis π3]2013=22013 cis 2013π3=22013(1+0i)=22013. \left[ 2 \text{ cis } \frac{ \pi}{3} \right]^{2013} = 2^{2013} \text{ cis } \frac{ 2013 \pi } { 3} = 2^{2013} ( - 1 + 0 i ) = - 2^{2013}.

 

2. Show that cos(5θ)=cos5θ10cos3θsin2θ+5cosθsin4θ \cos (5\theta) = cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta.

Applying De Moivre's Theorem for n=5n= 5 , we obtain

cos(5θ)+isin(5θ)=(cosθ+isinθ)5. \cos (5 \theta) + i \sin ( 5 \theta) = ( \cos \theta + i \sin \theta) ^ 5 .

Expand the RHS using the Binomial Theorem and compare real parts to obtain:

cos(5θ)=cos5θ10cos3θsin2θ+5cosθsin4θ. \cos ( 5 \theta) = \cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta.

Note: For an integer nn, we can express cos(nθ) \cos ( n \theta) solely in terms of cosθ \cos \theta by using the identity sin2θ=1cos2θ \sin^2 \theta = 1 - \cos^2 \theta. This is known as the Chebyshev polynomial of the first kind.

 

3. Evaluate sin(0θ)+sin(1θ)+sin(2θ)++sin(nθ) \sin (0\theta) + \sin (1 \theta) + \sin (2 \theta) + \ldots + \sin (n \theta) .

Applying De Moivre's Formula, this is equivalent to the imaginary part of

(cosθ+isinθ)0+(cosθ+isinθ)1+(cosθ+isinθ)2++(cosθ+isinθ)n. ( \cos \theta + i \sin \theta)^0 + ( \cos \theta + i \sin \theta)^1 + ( \cos \theta + i \sin \theta) ^2 + \ldots + ( \cos \theta + i \sin \theta)^n.

Interpreting this as a Geometric Progression, the sum is

(cosθ+isinθ)n+11(cosθ+isinθ)1, \frac{ (\cos \theta + i \sin \theta)^{n+1} -1} {( \cos \theta + i \sin \theta) - 1 },

as long as the ratio is not 1, which means θ2kπ \theta \neq 2k \pi . (Note that in this case, we get that each term sin(kθ) \sin (k\theta) is 0, and hence the sum is 0.)

Converting this to polar form, we obtain

ei(n+1)θ1eiθ1=ei(n+12)θei12θ×ei(n+12)θei(n+12)θei12θei12θ=ein2θ2isin[(n+12)θ]2isin(12θ). \frac{ e^{i (n+1) \theta} - 1 } { e^{i\theta} -1 } = \frac{ e^{ i \left( \frac{n+1}{2} \right)\theta} } {e^{i \frac{1}{2} \theta} } \times \frac{e^{ i \left( \frac{n+1}{2} \right)\theta} - e^{ - i \left( \frac{n+1}{2} \right)\theta} } { e^{ i \frac{1}{2} \theta} - e^{-i \frac{1}{2} \theta} } = e^{ i\frac{n}{2} \theta} \frac{2i \sin [ ( \frac{n+1}{2})\theta ] } { 2i \sin (\frac{1}{2} \theta)} .

Taking imaginary parts, we obtain

sin(n2θ)sin(n+12θ)sin(12θ). \frac{ \sin \left( \frac{n}{2} \theta \right) \sin \left( \frac{n+1}{2} \theta \right) } { \sin \left( \frac{1}{2} \theta \right) }.

Note by Calvin Lin
5 years, 3 months ago

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Find the sum of the following series : 1+2x+3x^2+4x^3+. . .. .nx^(n-1)

Bahram Dilfanian - 4 years, 1 month ago

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Derivative of geometric series.

A Brilliant Member - 11 months ago

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This was great! Question : How do we define cos(x) and sin(x) for complex numbers?

Roberto Nicolaides - 4 years, 5 months ago

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Do you know the Maclaurin series for sinθ \sin \theta about 0? That is one approach to take

sinx=xx33!+x55 \sin x = x - \frac {x^3}{3!}+ \frac{x^5}{5} - \ldots

This would be the analytic continuation, where we define it via (infinite) polynomials on the real line about 0, and then extend that to the complex plane around 0.

Calvin Lin Staff - 4 years, 5 months ago

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