De Moivre's Theorem

By David Marland

Introduction

Abraham De Moivre was a mathematician born in Champagne France on May 26, 1667. Since protestantism was not tolerated by the royal family in France at the age of 18 he decided to move to London. This gave him the opportunity to learn from highly respected Mathematicians in particular Isaac Newton, Edmond Halley, and James Stirling (O'Connor , 2004). He was unable to become a chair at any university and relied on his tutoring job to support him. Despite his challenges to keep out of poverty, he was a very competent Mathematician. One of his contributions to the field of Mathematics is the De Moivre's theorem. This formula connects complex numbers and trigonometry. In addition, it derives useful expressions for cos nx and sin nx in terms of cos x and sin x. (Daniels 2008)

[cos(θ) + isin(θ)]n = cos(nθ) + isin(nθ)

It should be noted that Euler, not De Moivre, wrote the theorem explicitly. (Nahin 1998).
First of all, we will prove De Moivre's theorem and then explore the many benefits of this formula.

Method of Induction

We can use method of induction to prove De Moivre's theorem.

Let n = 1

[cos(θ) + isin(θ)]1= cos(θ) + isin(θ)

Assume true for n = k, lets prove for n= k+1

[cos(θ) + isin(θ)]k+1 = [cos(θ) + isin(θ)]k[[cos(θ) + isin(θ)]]
[cos(θ) + isin(θ)]k+1 = [cos(kθ) + isin(kθ)][[cos(θ) + isin(θ)]]
[cos(θ) + isin(θ)]k+1 = [cos(kθ)cos(θ)-sin(kθ)sin(θ)] + i[sin(kθ)cos(θ) + cos(kθ)sin(θ)]

Compound angle formulae can simplify the expression.

[cos(θ) + isin(θ)]k+1 = cos((k+1)θ) + isin((k+1)θ)

We have proven for all positive integers that this theorem works but now we need to show for negative integers.


let n = -N where N is a positive integer.


[cos(θ) + isin(θ)]-N = [[cos(θ) + isin(θ)]N] -1

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Applying De Moivre's theorem


[cos(θ) + isin(θ)]-N = [cos(Nθ) + isin(Nθ)]-1
1 / [cos(Nθ) + isin(Nθ)]

Multiply the numerator and the denominator by the complex conjugage.


[cos(Nθ) - isin(Nθ)] / [[cos(Nθ) + isin(Nθ)][cos(Nθ) - isin(Nθ)]

[cos(Nθ) - isin(Nθ)] / [cos2(Nθ) + sin2(Nθ)]


This simplifies to
cos(Nθ) - isin(Nθ)
Using properties of even and odd functions, this can be written as
cos(-Nθ) + isin(-Nθ).
Hence, we have shown that

[cos(θ) + isin(θ)]-N = cos(-Nθ) + isin(-Nθ)

De Moivre's theorem is true for all integers. This is a one-to one-relationship. We will see in the section "finding the nth roots" that when 0<|n|<1 there exists more than one solution.

Application of De Moivre's Theorem

Evaluating (a + bi)n

Example (1 + i)10
Using Newton's binomial expansion is feasible but not a convenient method to compute it in the form x + yi. It will be even more burdensome if 1+ i is raised to higher powers.
Firstly, we should express 1 + i in the form r[cos(θ) + isin(θ)]
The magnitude of this complex number is √2. The complex number is in the first quadrant so the argument will be π⁄4.

1 + i = √2 [cos(π⁄4) + isin(π⁄4)]

Now let's evaluate (1 + i)10
[√2 [cos(π/4) + isin(π/4)]]10
√210 [cos(π⁄4) + isin(π⁄4)]]10
Applying De Moivre's theorem
√210 [cos(10π⁄4) + isin(10π⁄4)]]

We can see in this example that |(1+i)10 | = |1+i|10 and arg(1+i)10 = 10 arg(1+i).
This line with length √2, forms a counter-clockwise angle of 45 degrees from the positive x-axis. It has been rotated 10 times around the origin. Its magnitude has also increased by a power of 10.

Let z = cos(θ) + isin(θ), we can find cos(nθ) and sin(nθ) in terms of z

zn = [cos(θ) + isin(θ)]n

Applying De Moivre's theorem
zn = [cos(nθ) + isin(nθ)]
z-n = [cos(-nθ) + isin(-nθ)]
Using properties of even and odd functions.
z-n = [cos(nθ) - isin(nθ)]

Hence,
cos(nθ) = ½[zn + z-n]
sin(nθ) = 1⁄2i[zn - z-n]

Expressing cos(nθ) or sin(nθ) in terms of cos(θ) and sin(θ)

Example 1

Express cos(3θ) in terms of cos(θ)

cos(3θ) = Re[cos(3θ)+isin(3θ)]

Using De Moivre's theorem
cos(3θ) = Re[cos(θ)+isin(θ)]3

This can be expanded using Newton's binomial expansion.
cos(3θ) = Re[cos3(θ)+3icos2(θ)sin(θ)-3cos(θ)sin2(θ)- isin3(θ)]
cos(3θ) = cos3(θ)-3cos(θ)sin2(θ)

In terms of cos(θ)
cos(3θ) = cos3(θ)-3cos(θ)[1-cos2(θ)]

By looking at the imaginary part we can find an expression for sin(3θ)
Since sin(3θ) = Im[cos(3θ)+isin(3θ)]
Using De Moivre's theorem,
sin(3θ) = [cos(θ)+isin(θ)]3

This can be expanded using Newton's binomial expansion
sin(3θ) = Im[cos3(θ)+3icos2(θ)sin(θ)-3cos(θ)sin2(θ)-isin3(θ)]
sin(3θ) = 3cos2(θ)sin(θ)-sin3(θ)

In terms of sin(θ)
sin(3θ) = 3[1-sin2(θ)]sin(θ)-sin3(θ)

Example 2
Express cos3(θ) in terms of multiples of θ

cos(θ) = ½[z+z-1]
cos3(θ) = [1⁄2[z+z-1]]3
cos3(θ) = [1⁄8[z+z-1]]3

Using binomial expansion
cos3(θ) = 1⁄8[z3+3z1+3z-1+z-3]
cos3(θ) = 1⁄8[[z3+z-3]+3[z1+z-1]]
cos3(θ) = 1⁄4[cos(3θ)+3cos(θ)]

This application is particularly useful in integration, since sin(θ) or cos(θ) raised to a power means that we can now express these functions as a multiples of θ. As a result, we can easily apply method of substitution in order to integrate.
∫ cos3(θ) dθ = 1⁄4 ∫ [cos(3θ)+3cos(θ)]dθ

After applying substitution, we arrive at

∫ cos3(θ) dθ = 1⁄12cos(3θ)+3⁄4 sin(θ) + C

Finding the nth root

Although there are methods such as factor theorem, factorisation etc to solve zn = C (Mazur, 2003). De Moivre's theorem is a useful tool for finding the roots especially when n is large and C is a complex number.
Example - find the nth root of C where C is a real number

zn = C
z = R[cos(θ)+isin(θ)]
zn =[R[cos(θ)+isin(θ)]]n

Simplify then use Demoivre's Theorem


zn =Rn [cos(nθ)+isin(nθ)]] = C

C can be expressed as a complex number.
C = r [cos(α)+isin(α)]

Here r = C and α = 0 + 2jπ for j ∈ Z;
For simplicity, we will chose α = 0 but we can choose any j since the trigonometric functions are periodic every 2π.

C = C [cos(0)+isin(0)]]
Rn [cos(nθ)+isin(nθ)]] = C [cos(0)+isin(0)]]
Comparing the modulus

Rn = C
Since R has to be positive and real therefore R = C1/n.
Comparing the argument
nθ = 0 hence θ 1 = 0;
Cosine function is cyclical so therefore
nθ = 2π hence θ2 = 2π ⁄n
nθ = 4π hence θ3 = 4π ⁄n

If you continued this process, the solutions will simply repeat since θ is periodic every 2π.
z1 = R[cos(0)+isin(0)] = R
z2 = R[cos(2π ⁄n )+isin(2π ⁄n )]
zk = R[cos(2(k-1)π ⁄n )+isin(2(k-1)π ⁄n )]

According to the Fundamental Theorem of Calculus z n = C has n solutions (Carrera 1992) and examples below show that the solutions to z 3 = 1 and z 4 = 16 are all even distributed on the argand diagram.

Comparing the solutions to zn = C, zn = Ci, zn = -C, zn = -Ci


Example
Let's firstly compare zn = C, zn = iC, z3 = -C, z3 = -iC where C is a real number
On the right-hand of the equation we have multiplied by "i"
By the same procedure we can find the roots of zn = iC
The complex number "i" can be expressed in trigonometric form.
Here r = C and α = π⁄2

i = C [cos(π⁄2)+isin(π⁄2)]]
Rn [cos(3θ)+isin(3θ)]] = C [cos(π⁄2)+isin(π⁄2)]]

Comparing the modulus
Rn = C

R has to be the positive and real solution to R = C1/n.
Comparing the argument
nθ = π⁄2 hence θ 1 = π⁄2n;
Since the trigonometric functions have a period of 2π we can find the other solutions.
3θ = π⁄2+2π hence θ 2 = 5π⁄2n;
3θ = π⁄2+4π hence θ 3 = 9π⁄2n;

The unique solutions are therefore:
z1 = R[cos(π⁄2n)+isin(π⁄2n)]
z2 = R[cos(5π⁄2n)+isin(5π⁄2n)]
zk = R[cos(9π⁄2m)+isin(9π⁄2n)]

By multiplying the right-hand side by "i" the solutions have rotated by π⁄2n.

Continuing the investigation and finding the nth roots of -C the solutions would be:

z1 = R[cos(2π⁄2n))+isin(2π⁄2n)]
z2 = R[cos(6π⁄2n)+isin(6π⁄2n)]
zk = R[cos(10π⁄2n)+isin(10π⁄2n)]
Finally, the solutions to the nth root of -Ci would be:
z1 = R[cos(3π⁄2n))+isin(3π⁄2n)]
z2 = R[cos(7π⁄2n)+isin(7π⁄2n)]
zk = R[cos(11π⁄2n)+isin(11π⁄2n)]
It can be observed that every time we multiply the right-hand side of the equation zn = C by "i" the solutions are rotated by π⁄2n

Conclusion

De Moivre's theorem enables us to instantly see that |zn | = |z|n and also arg(zn) = n arg(z). We can avoid the process of having to expand brackets in order to express in modulus-argument form. It is revealed to us that raising a complex number by n rotates the original complex number around the origin n times and increases the magnitude by a power of n. We should be careful when n is a non-integer for example [cos(π)+isin(π)]1/3. De Moivre's theorem only provides one solution if apply the formula directly. The cube root of cos(π)+isin(π) can be expressed as a cubic where we are solving z 3 = cos(π)+isin(π). Since this is a polynomial of order three, the Fundamental Theorem of Algebra states that there will be three solutions (Carrera 1992). If we immediately applied De Moivre's Theorem, we will only find one of the solutions which is cos(π⁄3)+isin(π⁄3). Using the method in the previous section, the other solutions will be -1, cos(4/3π)+isin(4/3π). If we expand this theorem and use Euler's formula e iθ = cosθ + isin θ we can show that this occurs generally (Weisstein) . The limitation is that Euler's formula just shows one of the solutions when n is a non-integer.
As seen in previous examples De Moivre's theorem can be a useful tool in finding all the solutions to the nth root of a complex number. We can also spot the rotational symmetry between the solutions. The angle between the solutions will be 2π⁄n. For example, the cube root of 1 has an angle of 2π⁄3 between the solutions. Besides being a tool to find the roots of a polynomial, De Moivre's theorem also can be used to express powers of sin(θ) or cos(θ) as multiples of θ. As a result, we can use integration by substitution to integrate powers of sin(θ) or cos(θ).

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