MOIVRE'S THEOREM: PROOF AND SOLVED EXERCISES - MATHS - 2025

The theorem of Moivre applied algebra fundamental processes, such as powers and extracting roots in complex numbers. The theorem was stated by the renowned French mathematician Abraham de Moivre (1730), who associated complex numbers with trigonometry.

Abraham Moivre made this association through the expressions of the sine and cosine. This mathematician generated a kind of formula through which it is possible to raise a complex number z to the power n, which is a positive integer greater than or equal to 1.

What is Moivre's theorem?

Moivre's theorem states the following:

If we have a complex number in the polar form z = r _Ɵ, where r is the module of the complex number z, and the angle Ɵ is called the amplitude or argument of any complex number with 0 ≤ Ɵ ≤ 2π, to calculate its n– th power it will not be necessary to multiply it by itself n-times; that is, it is not necessary to make the following product:

Z ⁿ = z _* z _* z _{*… *} z = r _{Ɵ *} r _{Ɵ *} r _{Ɵ *… *} r _Ɵ n-times.

On the contrary, the theorem says that, when writing z in its trigonometric form, to calculate the nth power we proceed as follows:

If z = r (cos Ɵ + i _* sin Ɵ) then z ⁿ = r ⁿ (cos n * Ɵ + i _* sin n * Ɵ).

For example, if n = 2, then z ² = r ². If n = 3, then z ³ = z ² _* z. Also:

z ³ = r ² _* r = r ³.

In this way, the trigonometric ratios of the sine and cosine for multiples of an angle can be obtained, as long as the trigonometric ratios of the angle are known.

In the same way it can be used to find more precise and less confusing expressions for the n -th root of a complex number z, so that z ⁿ = 1.

To prove Moivre's theorem, the principle of mathematical induction is used: if an integer "a" has a property "P", and if for any integer "n" greater than "a" that has the property "P" It fulfills that n + 1 also has the property "P", then all integers greater than or equal to "a" have the property "P".

Demonstration

Thus, the proof of the theorem is done with the following steps:

Inductive base

It is first checked for n = 1.

Since z ¹ = (r (cos Ɵ + i _* sin Ɵ)) ¹ = r ¹ (cos Ɵ + i _* sin Ɵ) ¹ = r ¹, the theorem holds for n = 1.

Inductive hypothesis

The formula is assumed to be true for some positive integer, that is, n = k.

z ^k = (r (cos Ɵ + i _* sin Ɵ)) ^k = r ^k (cos k Ɵ + i _* sin k Ɵ).

Verification

It is proved to be true for n = k + 1.

Since z ^{k + 1} = z ^k _* z, then z ^{k + 1} = (r (cos Ɵ + i _* sin Ɵ)) ^{k + 1} = r ^k (cos kƟ + i _* sin kƟ) _* r (cos Ɵ + i _* senƟ).

Then the expressions are multiplied:

z ^{k + 1} = r ^{k + 1} ((cos kƟ) _* (cosƟ) + (cos kƟ) _* (i _* sinƟ) + (i _* sin kƟ) _* (cosƟ) + (i _* sin kƟ) _* (i _* senƟ)).

For a moment the factor r ^{k + 1} is ignored, and the common factor i is taken:

(cos kƟ) _* (cosƟ) + i (cos kƟ) _* (sinƟ) + i (sin kƟ) _* (cosƟ) + i ² (sin kƟ) _* (sinƟ).

Since i ² = -1, we substitute it in the expression and we get:

(cos kƟ) _* (cosƟ) + i (cos kƟ) _* (sinƟ) + i (sin kƟ) _* (cosƟ) - (sin kƟ) _* (sinƟ).

Now the real part and the imaginary part are ordered:

(cos kƟ) _* (cosƟ) - (sin kƟ) _* (sinƟ) + i.

To simplify the expression, the trigonometric identities of the sum of angles are applied for the cosine and sine, which are:

cos (A + B) = cos A _* cos B - sin A _* sin B.

sin (A + B) = sin A _* cos B - cos A _* cos B.

In this case, the variables are the angles Ɵ and kƟ. Applying the trigonometric identities, we have:

cos kƟ _* cosƟ - sin kƟ _* sinƟ = cos (kƟ + Ɵ)

sin kƟ _* cosƟ + cos kƟ _* sinƟ = sin (kƟ + Ɵ)

In this way, the expression is:

z ^{k + 1} = r ^{k + 1} (cos (kƟ + Ɵ) + i _* sin (kƟ + Ɵ))

z ^{k + 1} = r ^{k + 1} (cos + i _* sin).

Thus it could be shown that the result is true for n = k + 1. By the principle of mathematical induction, it is concluded that the result is true for all positive integers; that is, n ≥ 1.

Negative integer

Moivre's theorem is also applied when n ≤ 0. Let us consider a negative integer «n»; then "n" can be written as "-m", that is, n = -m, where "m" is a positive integer. Thus:

(cos Ɵ + i _* sin Ɵ) ⁿ = (cos Ɵ + i _* sin Ɵ) ^-m

To obtain the exponent «m» in a positive way, the expression is written inversely:

(cos Ɵ + i _* sin Ɵ) ⁿ = 1 ÷ (cos Ɵ + i _* sin Ɵ) ^m

(cos Ɵ + i _* sin Ɵ) ⁿ = 1 ÷ (cos mƟ + i _* sin mƟ)

Now, it is used that if z = a + b * i is a complex number, then 1 ÷ z = ab * i. Thus:

(cos Ɵ + i _* sin Ɵ) ⁿ = cos (mƟ) - i _* sin (mƟ).

Using that cos (x) = cos (-x) and that -sen (x) = sin (-x), we have:

(cos Ɵ + i _* sin Ɵ) ⁿ =

(cos Ɵ + i _* sin Ɵ) ⁿ = cos (- mƟ) + i _* sin (-mƟ)

(cos Ɵ + i _* sin Ɵ) ⁿ = cos (nƟ) - i _* sin (nƟ).

Thus, it can be said that the theorem applies to all integer values ​​of "n".

Solved exercises

Calculation of positive powers

One of the operations with complex numbers in their polar form is the multiplication by two of these; in that case the modules are multiplied and the arguments are added.

If you have two complex numbers z _{1 and} z ₂ and you want to calculate (z ₁ * z ₂) ², then you proceed as follows:

z ₁ z ₂ = *

The distributive property applies:

z ₁ z ₂ = r ₁ r ₂ (cos Ɵ _{1 *} cos Ɵ ₂ + i _* cos Ɵ _{1 *} i _* sin Ɵ ₂ + i _* sin Ɵ _{1 *} cos Ɵ ₂ + i ² _* sin Ɵ _{1 *} sin Ɵ ₂).

They are grouped, taking the term "i" as a common factor of the expressions:

z ₁ z ₂ = r ₁ r ₂

Since i ² = -1, it is substituted in the expression:

z ₁ z ₂ = r ₁ r ₂

The real terms are regrouped with real, and imaginary with imaginary:

z ₁ z ₂ = r ₁ r ₂

Finally, the trigonometric properties apply:

z ₁ z ₂ = r ₁ r ₂.

In conclusion:

(z ₁ * z ₂) ² = (r ₁ r ₂) ²

= r ₁ ² r ₂ ².

Exercise 1

Write the complex number in polar form if z = - 2 -2i. Then, using Moivre's theorem, calculate z ⁴.

Solution

The complex number z = -2 -2i is expressed in the rectangular form z = a + bi, where:

a = -2.

b = -2.

Knowing that the polar form is z = r (cos Ɵ + i _* sin Ɵ), we need to determine the value of the modulus "r" and the value of the argument "Ɵ". Since r = √ (a² + b²), the given values ​​are substituted:

r = √ (a² + b²) = √ ((- 2) ² + (- 2) ²)

= √ (4 + 4)

= √ (8)

= √ (4 * 2)

= 2√2.

Then, to determine the value of «Ɵ», the rectangular shape of this is applied, which is given by the formula:

tan Ɵ = b ÷ a

tan Ɵ = (-2) ÷ (-2) = 1.

Since tan (Ɵ) = 1 and we have a <0, then we have:

Ɵ = arctan (1) + Π.

= Π / 4 + Π

= 5Π / 4.

As the value of «r» and «Ɵ» has already been obtained, the complex number z = -2 -2i can be expressed in polar form by substituting the values:

z = 2√2 (cos (5Π / 4) + i _* sin (5Π / 4)).

Now we use Moivre's theorem to calculate z ⁴:

z ⁴ = 2√2 (cos (5Π / 4) + i _* sin (5Π / 4)) ⁴

= 32 (cos (5Π) + i _* sin (5Π)).

Exercise 2

Find the product of the complex numbers by expressing it in polar form:

z1 = 4 (cos 50 ^o + i _* sin 50 ^o)

z2 = 7 (cos 100 ^o + i _* sin 100 ^o).

Then calculate (z1 * z2) ².

Solution

First the product of the given numbers is formed:

z ₁ z ₂ = *

Then the modules are multiplied with each other, and the arguments are added:

z ₁ z ₂ = (4 _* 7) _*

The expression is simplified:

z ₁ z ₂ = 28 _* (cos 150 ^o + (i _* sin 150 ^o).

Finally, Moivre's theorem applies:

(z1 * z2) ² = (28 _* (cos 150 ^o + (i _* sin 150 ^o)) ² = 784 (cos 300 ^o + (i _* sin 300 ^o)).

Calculation of negative powers

To divide two complex numbers z _{1 and} z ₂ in their polar form, the modulus is divided and the arguments are subtracted. Thus, the quotient is z ₁ ÷ z ₂ and is expressed as follows:

z ₁ ÷ z ₂ = r1 / r2 ().

As in the previous case, if we want to calculate (z1 ÷ z2) ³, the division is carried out first and then the Moivre theorem is used.

Exercise 3

Dices:

z1 = 12 (cos (3π / 4) + i * sin (3π / 4)), z2 = 4 (cos (π / 4) + i * sin (π / 4)), calculate (z1 ÷ z2) ³.

Solution

Following the steps described above it can be concluded that:

(z1 ÷ z2) ³ = ((12/4) (cos (3π / 4 - π / 4) + i * sin (3π / 4 - π / 4))) ³

= (3 (cos (π / 2) + i * sin (π / 2))) ³

= 27 (cos (3π / 2) + i * sin (3π / 2)).

References

1. Arthur Goodman, LH (1996). Algebra and trigonometry with analytical geometry. Pearson Education.
2. Croucher, M. (nd). From Moivre's Theorem for Trig Identities. Wolfram Demonstrations Project.
3. Hazewinkel, M. (2001). Encyclopaedia of Mathematics.
4. Max Peters, WL (1972). Algebra and Trigonometry.
5. Pérez, CD (2010). Pearson Education.
6. Stanley, G. (nd). Linear algebra. Graw-Hill.
7. , M. (1997). Precalculation. Pearson Education.