Powers of Complex Numbers in Polar Form
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.
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)=cos3θ−3cosθsin2θ
| $\cos\left(3\theta\right)+i\sin\left(3\theta\right)$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+3icos2θsinθ−3cosθsin2θ−isin3θ | |
| $=$= | $\cos^3\theta-3\cos\theta\sin^2\theta+i\left(3\cos^2\theta\sin\theta-\sin^3\theta\right)$cos3θ−3cosθsin2θ+i(3cos2θsinθ−sin3θ) |
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)=cos3θ−3cosθsin2θ
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.
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).
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.
Rewrite $\left(\cos\alpha+i\sin\alpha\right)^2$(cosα+isinα)2 using De Moivre's Theorem.
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.
Hence write the expression for $\cos2\alpha$cos2α.
Hence write the expression for $\sin2\alpha$sin2α.