NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Powers of Complex Numbers in Polar Form
Lesson

Abraham de Moivre (1667 – 1754) was a French mathematician who moved to England. It was here where he associated with Newton and Halley and became a private teacher of mathematics. One of his most predominant contributions was to the field of complex numbers. De Moivre’s theorem relates to finding powers of complex numbers.

Before we learn the shortcut – it is important to appreciate the pattern that develops through consecutive applications of multiplication. Start with the activity below.

Activity

Using the multiplication rule for complex numbers in polar form, where $\left(r_1,\theta_1\right)\times\left(r_2,\theta_2\right)=\left(r_1r_2,\theta_1+\theta_2\right)$(r1,θ1)×(r2,θ2)=(r1r2,θ1+θ2) to

a) investigate $\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)^n$(cosπ6+isinπ6)n when $n=0,1,2,3,...,12$n=0,1,2,3,...,12

b) investigate $\left(3\cos\frac{\pi}{6}+3i\sin\frac{\pi}{6}\right)^n$(3cosπ6+3isinπ6)n when $n=0,1,2,3,...,6$n=0,1,2,3,...,6

What is the pattern you notice?

 

What you should have discovered is that $\left(\cos\theta+i\sin\theta\right)^n=\cos\left(n\theta\right)+i\sin\left(n\theta\right)$(cosθ+isinθ)n=cos(nθ)+isin(nθ)

In modulus argument form, this is $\left(r,\theta\right)^n=\left(r^n,n\theta\right)$(r,θ)n=(rn,nθ)

This is de Moivre's theorem and this is true for any rational number $n$n.  

There are many applications of de Moivre's theorem, one of them is in simplification of complex algebraic terms and another in the proof of trigonometric identities. 

 

Example 1

Prove that $\cos\left(3\theta\right)=\cos^3\theta-3\cos\theta\sin^2\theta$cos(3θ)=cos3θ3cosθsin2θ

$\cos\left(3\theta\right)+i\sin\left(3\theta\right)$cos(3θ)+isin(3θ) $=$= $\left(\cos\theta+i\sin\theta\right)^3$(cosθ+isinθ)3
  $=$= $\cos^3\theta+3\cos^2\theta\left(i\sin\theta\right)+3\cos\theta\left(i\sin\theta\right)^2+\left(i\sin\theta\right)^3$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+3icos2θsinθ3cosθsin2θisin3θ 
  $=$= $\cos^3\theta-3\cos\theta\sin^2\theta+i\left(3\cos^2\theta\sin\theta-\sin^3\theta\right)$cos3θ3cosθsin2θ+i(3cos2θsinθsin3θ)

Comparing real parts of the equation we get that $\cos\left(3\theta\right)=\cos^3\theta-3\cos\theta\sin^2\theta$cos(3θ)=cos3θ3cosθsin2θ

 

Example 2

Simplify the following expression $\frac{\cos2\theta+i\sin2\theta}{\cos3\theta+i\sin3\theta}$cos2θ+isin2θcos3θ+isin3θ

$\frac{\cos2\theta+i\sin2\theta}{\cos3\theta+i\sin3\theta}$cos2θ+isin2θcos3θ+isin3θ $=$= $\frac{\left(\cos\theta+i\sin\theta\right)^2}{\left(\cos\theta+i\sin\theta\right)^3}$(cosθ+isinθ)2(cosθ+isinθ)3
  $=$= $\frac{1}{\left(\cos\theta+i\sin\theta\right)^1}$1(cosθ+isinθ)1
  $=$= $\frac{1}{\cos\theta+i\sin\theta}$1cosθ+isinθ
  $=$= $\left(\cos\theta+i\sin\theta\right)^{-1}$(cosθ+isinθ)1
  $=$= $\left[\cos\left(-\theta\right)+i\sin\left(-\theta\right)\right]$[cos(θ)+isin(θ)]
  $=$= $\cos\left(\theta\right)-i\sin\left(\theta\right)$cos(θ)isin(θ)

 

Worked examples

Question 1

Find $\left(\cos40^\circ+i\sin40^\circ\right)^9$(cos40°+isin40°)9.

Give your answer in rectangular form.

Question 2

Find $\left(2\left(\cos110^\circ+i\sin110^\circ\right)\right)^3$(2(cos110°+isin110°))3.

Give your answer in rectangular form.

Question 3

Consider the expression $\left(\sqrt{3}+i\right)^8$(3+i)8.

  1. When raising a complex number to a power, we can use De Moivre's Theorem to evaluate the expression, but the complex number has to be in polar form.

    First, rewrite $\sqrt{3}+1i$3+1i in polar form, $r\left(\cos\theta+i\sin\theta\right)$r(cosθ+isinθ).

  2. Now evaluate $\left(\sqrt{3}+i\right)^8$(3+i)8 using De Moivre's Theorem, expressing this complex number in polar form.

Question 4

Answer the following.

  1. Rewrite $\left(\cos\alpha+i\sin\alpha\right)^2$(cosα+isinα)2 using De Moivre's Theorem.

  2. Rewrite $\left(\cos\alpha+i\sin\alpha\right)^2$(cosα+isinα)2 by expanding the brackets.

    Give your answer in the form $a+bi$a+bi.

  3. Hence write the expression for $\cos2\alpha$cos2α.

  4. Hence write the expression for $\sin2\alpha$sin2α.

 

Outcomes

M8-9

Manipulate complex numbers and present them graphically

91577

Apply the algebra of complex numbers in solving problems

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