Euler’s Formula for Complex Analysis and Polyhedron with Examples (2023)

Introduced by Leonhard Euler, a Swiss mathematician, Euler’s formula can be defined by two different ways:

1. For complex analysis: Euler’s formula is used in complex analysis to find the solution of complex exponential functions. It can also be termed as Euler’s identity and establishes a relationship between trigonometric functions and exponential functions.
2. For Polyhedra: For a non self-intersecting polyhedra, the Euler’s formula establishes a relation between the number of faces, vertices, and edges of the polyhedra.

In this maths article, we shall learn about Euler’s formula for complex analysis and polyhedra in detail. We shall also derive Euler’s formula using different methods. Also, we will check the applications of Euler’s formula along with solved examples for better understanding of the concept.

What is Euler’s Formula?

Euler’s formula is a theorem which states that:

\(e^{ix}=\cos \theta +i\sin \theta \)

Here, ‘x’ is a real number.

‘e’ is the base of the natural log.

And ‘i’ is the imaginary number.

Euler’s formula sets up a fundamental relationship between trigonometric functions and exponential functions.

Or,

It can also be considered as a way to bridge the two representations of the same unit complex numbers in a complex plane.

Derivative of Euler’s Formula

We can derive Euler’s formula in three different ways. The first is to drive the formula using the power series. In this type of derivative exponential, sine and cosine functions are expanded as power series.

In the second tupe, we derive Euler’s formula using Calculus. Here, both the sides of the equation are treated as functions and are then differentiated.

And finally, the third derivation uses polar coordinates on the complex plane to find the values of r and \(\theta\).

Derivation 1: Using Power Series

To begin with, let us assume that power series expansion of sin z, cos z, and \(e^z\) are absolutely convergent everywhere.

For a complex variable ‘z’, the power series expansion of \(e^z\) is

\(e^z=1+\frac{z}{1!}+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+…\)

Let us take ‘z’ as ‘ix’ (here x is any complex number).

As ‘z’ gets raised to increasing power, ‘i’ also gets raised to increasing power. Some of the powers of ‘i’ are:

\(i^0=1, i^1=i, i^2= -1, i^3= i.i^2 = -i, i^4=i^2.i^2 = 1,i^5= i.i^4=i, i^6=i.i^5 = i^2 = -1, i^7 = i.i^6 = -i\)

With z = ix the expansion \(e^z\) becomes:

\(e^{ix}=1+\frac{ix}{1!}+\frac{\left(ix\right)^2}{2!}+\frac{\left(ix\right)^3}{3!}+\frac{\left(ix\right)^4}{4!}+…\)

Using the powers of ‘i’, we get:

\(e^{ix}=1+ix-\frac{x^2}{2!}-\frac{ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}-\frac{x^6}{6!}-\frac{ix^7}{7!}+\frac{x^8}{8!}+…\)

As the power series of expansion of \(e^z\) is absolutely convergent, let us rearrange its terms without changing the values. Let us group real and imaginary terms:

\(e^{ix}=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}+…\right)+i\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+…\right)\)

Let us consider power series of sine and cosine:

Power series of cos x is:

Cos x = \(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}+…\)

And the power series of sin x is:

Sin x = \(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+…\)

Or, we can write:

\(e^{ix}=\cos x+i\sin x/latex]

Which is the statement for Euler’s formula.

Derivation 2: Using Calculus

Let us consider \(\)e^{ix}\) and cos x + i sin x as functions of x, and then differentiate them to find the common property between them.

But before that let us assume that \(e^z\), cos x, and sin x are defined and differentiable for real numbers x and complex numbers z.

Let \(f_1(x) = e^{ix}\) and \(f_2(x) = cos x +i sin x\)

Using chain rule:

\(f_1^{\prime}\left(x\right)=ie^{ix}=if_1\left(x\right)/latex] \(\)f_2^{\prime}\left(x\right)=-\sin x+i\cos x=if_2\left(x\right)\)

We can say that \(f_1(x)\) and \(f_2(x)\) satisfy f’(x) = if(x).

Consider \(\frac{f_1\left(x\right)}{f_2\left(x\right)}\) which is well defined for all x.

Using quotient rule:

\(\left(\frac{f_1}{f_2}\right)^{\prime}\left(x\right)=\frac{f_1^{\prime}\left(x\right)f_2\left(x\right)-f_1\left(x\right)f_2^{\prime}\left(x\right)}{\left[f_2\left(x\right)\right]^2}\)

\(=\frac{if_1\left(x\right)f_2\left(x\right)-f_1\left(x\right)if_2\left(x\right)}{\left[f_2\left(x\right)\right]^2}\) = 0

As derivative of \(\left(\frac{f_1}{f_2}\right)\left(x\right) = 0\), that implies \(\frac{f_1}{f_2}\) must be constant.

To find this constant, let us start with x = 0,

\(\left(\frac{f_1}{f_2}\right)\left(0\right)=\frac{e^{i0}}{\cos 0+i\sin 0}=1\)

Or,

\(\left(\frac{f_1}{f_2}\right)\left(x\right) = \frac{e^{ix}}{\cos x+i\sin x}=1\)

Moving cos x + i sin x to right, we get:

\(e^{ix}=\cos x+i\sin x\)

Hence, we get the statement of Euler’s formula.

Derivation 3: Using Polar Coordinates

To prove Euler’s formula using polar coordinates we need to treat exponentials as complex numbers under polar coordinates.

We know that all the non-zero complex numbers can be uniquely expressed in polar coordinates.

Any number \(e^{ix}\) (where x belongs to R) can be expressed as:

\(e^{ix}=r\left(\cos \theta +i\sin \theta \right)\)

Here, \(\theta\) is the principal angle from the positive real axis, and r is the radius. (Also, we know that, x = 0, then LHS = 1, r and \(\theta\) satisfy initial conditions of r(0) = 1 and \(\theta(0) = 0\).

Let us differentiate both sides of the equation:

\(ie^{ix}=\frac{dr}{dx}\left(\cos \theta +i\sin \theta \right)+r\left(-\sin \theta +i\cos \theta \right)\frac{d\theta }{dx}\)

Replace \(e^{ix}\) by \(r\left(\cos \theta +i\sin \theta \right)\) to get an expression in ‘r’ and \(\theta\).

\(ir\left(\cos \theta +i\sin \theta \right)=\left(\cos \theta +i\sin \theta \right)\frac{dr}{dx}+r\left(-\sin \theta +i\cos \theta \right)\frac{d\theta }{dx}\)

\(r\left(i\cos \theta -\sin \theta \right)=\left(\cos \theta +i\sin \theta \right)\frac{dr}{dx}+r\left(-\sin \theta +i\cos \theta \right)\frac{d\theta }{dx}\)

Equating imaginary and real parts:

\(r\cos \theta = \sin \theta \frac{dr}{dx} + r\cos \theta \frac{d\theta }{dx}\)

\(-r\sin \theta = \cos \theta \frac{dr}{dx} – r\sin \theta \frac{d\theta }{dx}\)

Let, \(\frac{dr}{dx} = \alpha\), and \(\frac{d\theta }{dx}=\beta\)

Therefore,

\(r\cos \theta = (\sin \theta) \alpha + (r\cos \theta) \beta\)…(1)

\(-r\sin \theta = (\cos \theta) \alpha – (r\sin \theta) \beta\)…(2)

Multiply (1) by \(cos\theta\) and (2) by \(sin\theta \)

\(r\cos ^2\theta =(\sin \theta \cos \theta )\alpha +(r\cos ^2\theta )\beta \)…(3)

\(-r\sin ^2\theta =(\sin \theta \cos \theta )\alpha -(r\sin ^2\theta )\beta \)…(4)

Eliminate \(\theta \) by (3) – (4):

\(r\left(\sin ^2\theta +\cos ^2\theta \right)=r\left(\sin ^2\theta +\cos ^2\theta \right)\beta \)

As \(\sin ^2\theta +\cos ^2\theta = 1\)

Therefore, \(r=r\beta /latex]

As r > 0 for all ‘x’ that means \(\)\beta = 1 or, \(\)\frac{d\theta }{dx}=1\)

Again substitute the value of \(\beta \) in (1)and (2) and compare:

\(0=\left(\sin \theta \right)\alpha \)

\(0=\left(\cos \theta \right)\alpha \)

That implies \(\alpha = 0\) or \(\frac{dr}{dx}=0\)

As \(\frac{dr}{dx}=0\), that implies ‘r’ is a constant and as \(\frac{d\theta }{dx}=1\), \(\theta =x+C\), where C is constant.

As r satisfies the condition \(r(\theta) = 1\), therefore r = 1.

And \(\theta\) satisfies the condition \(\theta(0) = 0\), so, C = 0, i.e. \(\theta = x\)

As ‘r’ and \(\theta\) are defined, therefore our original equation becomes:

\(e^{ix}=r\left(\cos \theta +i\sin \theta \right)\) = cos x + sin x

Hence, we get the statement for Euler’s formula.

Euler’s Formula for Complex Analysis

In complex analysis Euler’s formula is used to set up a relationship between trigonometric functions and exponential functions.

For any real number ‘x’, Euler’s formula is defined as:

\(e^{ix}=\cos \theta +i\sin \theta /latex]

Here, cos x, and sin x are trigonometric functions, ‘i’ is an imaginary number, and ‘e’ is the base of a natural log.

In the complex plane, the formula can be interpreted as a unit complex function \(\)e^{i\theta}\). Here \(\theta\) is a real number and is measured in radians.

Proof:

Let us prove the above result using the expansion series of \(e^x\).

\(e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+…\infty \)

Let us suppose that the expansion is true even if the value of ‘x’ is non-real.

Also, let \(x = i\theta\)

\(e^{i\theta }=1+\frac{i\theta }{1!}+\frac{\left(i\theta \right)^2}{2!}+\frac{\left(i\theta \right)^3}{3!}+\frac{\left(i\theta \right)^4}{4!}+…\infty \)

\(e^{i\theta }=1+\frac{i\theta }{1!}-\frac{\theta ^2}{2!}-\frac{i\theta ^3}{3!}+\frac{\theta ^4}{4!}+…\infty \) (as \( i^2 = -1\))

\(e^{i\theta }=\left(1-\frac{\theta ^2}{2!}+\frac{\theta ^4}{4!}+…\infty \right)+i\left(\frac{\theta }{1!}-\frac{\theta ^3}{3!}+\frac{\theta ^5}{5!}…\infty \right)\)

The above two series are Taylor’s expansion series for \(cos\theta \) and \(sin\theta \).

Therefore;

\(e^{ix}=\cos x+i\sin x\)

Euler’s Identity

We know that by Euler’s Formula \(e^{ix}=\cos x+i\sin x\). The formula can be expressed as identity, where \( is replaced by \(\)\pi \).

\(e^{i\pi}=\cos \pi+i\sin \pi\)

\(e^{i\pi}=-1 + i(0) \) (as \(cos \pi = -1 and sin\pi = 0\))

\(e^{i\pi} = -1\)

Or,

\(e^{i\pi} + 1=0\)

Euler’s Formula for Polyhedra

Geometrically, a polyhedron is a three-dimensional shape that has flat surfaces and straight edges. Some of the examples of polyhedra are cube, prism, cuboid, and pyramid.

The faces, edges and vertices of a polyhedron that do not self intersect are related in a certain manner.

According to Euler’s formula for polyhedra, the number of faces and vertices of a polyhedron taken together are two more than the number of edges. We can write this as:

F + V – E =2

Here, F is the number of Faces

V is the number of vertices

And E is the number of edges of a polyhedron.

Proof:

A graph is formed by drawing dots and lines. A planar graph is obtained when no lines or edges cross each other. Using the same statement a cube can be plotted as a planar graph by projecting vertices and edges on a plane. Euler’s formula graph theory states that number of dots – number of lines + number of regions cut by plane = 2.

Verification of Euler’s Formula

We can use Euler’s formula for different solid shapes and complex polyhedra.

Let us take an example of a simple polyhedron like square pyramid and triangular prism to verify Euler’s formula.

Example 1: Square Pyramid:

Number of faces = 5

Number of vertices = 5

Number of edges = 8

Therefore; F + V – E = 5 + 5 – 8 = 2

So, the Euler’s formula holds true.

Example 2: Triangular Prism

Number of faces = 5

Number of vertices = 6

Number of edges = 9

Therefore; F + V – E = 5 + 6 – 9 = 2

So, the Euler’s formula holds true.

Applications of Euler’s Formula

As Euler’s formula is an important mathematical equation, it has some interesting applications. These are:

• It is used in Euler’s identity.
• For the exponential form of complex numbers.
• As an alternative definition of trigonometric and hyperbolic functions.
• For generalization of exponential and logarithmic functions to complex numbers.
• As an alternative proof of de Moivre’s theorem and trigonometric additive identities.

Euler’s Formula Explanation

In all there are five platonic solids for which we can use Euler’s formula. These solids are:

Cube, Tetrahedron, Octahedron, Dodecahedron, and Icosahedron

The following table verifies the Euler’s formula for complex polyhedron:

Euler’s Formula Solved Examples

Que 1: A polyhedron has 12 vertices and 30 edges. Find the number of faces of the polyhedron.

Ans 1: Given that:

Number of vertices of a polyhedron = 12

Number of edges of a polyhedron = 30

We know that:

By Euler’s formula;

F + V – E = 2

Therefore,

F + 12 – 30 = 2

F = 2 – 12 + 30 = 20

Therefore, the faces of the given polyhedron are 20.

Que 2: Express \(3e^{5i}\) in (a + ib) using Euler’s formula.

Ans 2: Given that:

\(\theta = 5\)

Using Euler’s formula:

\(e^{i\theta }=\cos \theta +i\sin \theta \)

\(e^{5i} = cos 5 + i sin 5\) = 0.284 + i(-0.959) = 0.284 – 0.959i

So,

\(3e^{5i}= 0.852 – 2.877i

Therefore, we can express \(\)3e^{5i}\) in (a + ib) as 0.852 – 2.77i.

Que 3: Express \(e^{i\left(\frac{\pi }{2}\right)}\) in (a + ib) using Euler’s formula.

Ans 3: Given that:

\(\theta = \frac{\pi}{2}\)

Using Euler’s formula:

\(e^{i\theta }=\cos \theta +i\sin \theta \)

\(e^{i\left(\frac{\pi }{2}\right)} = cos(\frac{\pi}{2})+ i sin (\frac{\pi}{2})\)= 0 + \(i\times 1\) = i

Therefore, we can express \(e^{i\left(\frac{\pi }{2}\right)}\) in (a + ib) as i.

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