Eulers Formula and De Moivres Theorem in Complex Numbers

We know about complex numbers(z). They are of the form z=a+ib, where a and b are real numbers and 'i' is the solution of equation x²=-1. No real number can satisfy this equation hence its solution that is 'i' is called an imaginary number. When a  complex exponential is written, it is written as e^iθ.

Euler's formula explains the relationship between complex exponentials and trigonometric functions.

DeMoivers’ theorem is also a theorem used for complex numbers. This theorem is used to raise complex numbers to different powers.

State Euler's Theorem

Euler’s law states that ‘For any real number x, e^ix = cos x + i sin x.

where,e=base of natural logarithm

           i=imaginary unit

          x=angle in radians

This complex exponential function is sometimes denoted cis x ("cosine plus i sine"). The formula is still valid if x is a complex number.

Let z be a non zero complex number; we can write z in the polar form as,

z = r(cos θ + i sin θ) = r e^iθ, where r is the modulus and θ is argument of z.

Multiplying a complex number z with e^iα gives, zei^α = re^iθ × ei^α = rei^(α + θ).The resulting complex number re^i(α+θ) will have the same modulus r and argument (α+θ).

Euler's Identity 

When x=π Euler’s formula evaluates to e^iπ+1=0, which is known as Euler's Identity.

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Euler's Formula 

Euler's Formula For Cube

Euler's formula is related to the Faces, Edges and vertices of any polyhedron. 

Euler's formula for a cube says that in a cube, the number of vertices minus the number of edges plus the number of faces results in two.

It can be written as 

                              V-E+F=2

Where, V=number of vertices

           E=number of edges

           F=number of faces

It can be proven as,

In a cube, the number of vertices = 8

                number of edges= 12

                number of faces= 6

Putting values into the formula, V-E+F=8-12+6

                                                     =2

Hence proved.

De Moiver's Theorem

State De Moiver's Theorem

It states that for any integer n,

(cos θ + i sin θ)^n = cos (nθ) + i sin (nθ)

We can prove this easily using Euler’s formula as given below,

We know that, (cos θ + i sin θ) = e^iθ

(cos θ + i sin θ)^n = e^i(nθ)

Therefore,

e^i(nθ) = cos (nθ) + i sin (nθ)

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nth Roots of Unity

If any complex number satisfies the equation zn = 1, it is known as nth root of unity.

An equation of degree n will have n roots as said by the fundamental theory of algebra, there are n values of z which satisfies zn = 1.

To find the values of z, we can write,

1 = cos (2kπ) + i sin (2kπ), —(1) where k can be any integer.

We have,

z^n = 1

z = 1^(1/n)

From (1),

z = [cos (2kπ) + i sin (2kπ)]^(1/n)

By De Moivre’s theorem,

z = [cos (2kπ/n) + i sin (2kπ/n)], where k = 0,1,2,3,……..,n−1

For example; if n = 3, then k = 0,1,2

We know that, z = cos (2kπ/n) + i sin (2kπ/n) = e^i(2kπ/n)

Let ω = cos (2πn) +i sin (2πn) = e^i(2πn)

nth roots of unity are found by,

When k = 0; z = 1

k = 1; z = ω

k = 2; z = ω²

k = n; z = ω^{n − 1}

Therefore, nth roots of unity are 1,ω,ω^2,ω³,…….,ω^{n − 1}

Sum of nth roots of unity is,1 + ω + ω² + ω³ + ⋯ + ω^{n − 1}It is geometric series having first term 1 and common ratio ω.By using sum of n terms of a G.P,1 + ω + ω² + ω³ + ⋯ + ω^{n − 1} = 1 − ω^{n1 − ω}Since ω is nth root of unity, ωn = 1Therefore, 1 + ω + ω² + ω³ + ⋯ + ω^{n − 1} = 0

Cube Roots of Unity:

We know that nth roots of unity are 1,ω,ω2,ω3,…….,ωn − 1.

Therefore, cube roots of unity are 1,ω,ω2 where,

ω = cos (2π/3) + i sin (2π/3) = −1 + √3 i2

ω² = cos(4π/3) + i sin (4π/3) = −1 − √3 i2

Sum of the cube roots of the unity,

1 + ω + ω² = 0

Product of cube roots of the unity,

1 × ω × ω² = ω³ = 1

De Moiver's Theorem Example 

If z = (cosθ + i sinθ ) , show that z^n + 1/ z^n = 2 cos nθ and z^n – [1/ z^n] = 2i sin nθ .

Solution

Let z = (cosθ + i sinθ ) . 

By de Moivre’s theorem ,

z^n = (cosθ + i sinθ )^n = cos nθ + i sin nθ

1/z^n=z^(-n)=cos nθ - i sin nθ

=> z^n+1/z^n = (cos nθ + i sin nθ)+(cos nθ - i sin nθ)

=> z^n+1/z^n = 2cosnθ

Also,=> z^n-1/z^n = (cos nθ + i sin nθ)-(cos nθ - i sin nθ)

=> z^n-1/z^n = 2i sin nθ