The Jewel In The Crown Of Mathematics 

In mathematics, imaginary quantities are unsolvable, unreal and impractical yet these are expressible geometrically, thanks to Swiss mathematician Leonhard Euler who made this possible. For example ‘i’which is square root of -1, can be expressed as a unit length along Y axis and 1, is expressed as unit length along X axis. A complex quantity a + i.b denotes to a point that corresponds to length ‘a’ along X axis and ‘b’ along Y axis. How it is made possible is a subject matter of discussion along with independent derivation of Euler’s formula and its related issues.

Euler Formula And Identity

Leonhard Euler stated that natural number ‘e’ when is raised to power imaginary number, is expressible and solvable. This caused flutter in mathematicians fraternity. It was known that natural number raised to real quantity is real and solvable but how power of an imaginary quantity can result in real and imaginary was baffling. Euler not only limited himself to expansion of imaginary power to real quantity resulting in real and imaginary quantity but also proved that imaginary power of natural number can yield pure real quantity also. This was later known as Euler’s identity and is mathematically written as

e^i.pi + 1 = 0 or

e^i.pi = – 1 where sign ^ denotes raised to power.

Euler’s formula can be written mathematically as

e^i.x = cos x + i.sin x.

Here we will derive this Formula independently, even Binomial expansion and derivative of a function will be derived ab initio.

Binomial Expansion

We know, by simple multiplication,

(x + a)^2 = x^2 + 2.a.x + a^2 = 2C2.x^2 + 2C1.x.a^1 +2C0.a^2.


(x + a)^3 = x^3 + 3.a.x^2 + 3.a^2.x + a^3 = 3C3.x^3 + 3C2.x^2.a + 3.C2.x.a^2 + 3C3.a^3.

By mathematical induction,

(x+ a)^n = nCn x^n + nCn-1. X^n-1. a + nCn-2.x^n-2.a^2 +……nCr.x^n-r.a^r…………..nC0. a^n

where nCr = n!/(n-r)!.r! and sign ! denoted factorial.

On putting, a =1 in above expansion,

(1+ x)^n = 1 + n.x + [n(n-1)]/2!.x^2 + [n(n-1)(n-2)]/3!.x^3+…………… [n(n-1)…(n-r)/r!.x^n-r +……………..(1).

Equation (1) is a binomial expansion of (1+ x)^n. If x is less than 1, and n is defined then infinite term tends to zero.

Derivative of a function

Derivative of a function f(x) is defined as

 f’(x) = [f(x + delta x) – f(x)]/delta x where delta x is infinitesimal tending to zero.

Let us find derivative of tan x, since it will be used in derivation of Euler Formula. Applying above formula,

derivative of tan x or d/dx of tan x

= limit delta x tends to 0, [tan (x+ delta x) – tan x]/delta x,

We know tan (x+ delta x) = (tan x + tan delta x)/(1- tan x. tan delta x). Putting this value, we get

Derivative of tan x = [tan x + tan delta x – tan x + tan^2 x. tan delta x]/[(1- tan x. tan delta x).delta x]

= limit delta x tends to zero, (tan delta x/ delta x).[(1+ tan^2 x)/(1- tan x. tan delta.x)]

= 1.(1+ tan^2x)/(1- tan x. 0) = sec^2 x,

Since (tan delta x)/delta x = 1 and tan delta x = 0 when delta x tends to 0.

Therefore, derivative or d/dx of tan x = sec^2 x………….(2).

Expansion of tan inverse x in power series

Let tan inverse x = y, then

tan y = x.

On differentiating both sides,

Sec^2 y. dy/dx = 1 from equation (2).

Or dy/dx = 1/ sec^2 y = 1/(1+ tan^2 y) = 1/(1+ x^2).

Expanding 1/(1+ x^2) in binomial expansion in accordance with equation (1),

(1 + x^2)^-1 = 1 – x^2 + (-1)(-2)/2!.x^4 + (-1)(-2).(-3)/3!+……….. or

(1+ x^2)^-1 = 1- x^2 + x^4 – x-6 + x^8……………….

On integrating both sides, we get

y = tan inverse x = c + x – x^3/3 + x^5/5 -x^7/7+………..

At x = 0, tan inverse 0 = 0.

Therefore, 0 = c.

Tan inverse x = x – x^3/3 + x^5/5 -x^7/7+…………………..

Since tangent of an angle repeats its value after an angle of 2pi, On putting tan inverse x as y + 2k.pi where k is an integer,

 2k.pi+ y= tan ( y) – tan^3 ( y) /3 + tan^5 (y) /5+…………………..(3).

Expansion of log(1+ x)

Similarly 1/(1+x) can be expanded by Binomial theorem in accordance with equation (1) as

(1+x)-1 = 1-x + x^2 -x^3 + x^4 – x^5………………..

On integrating both sides with respect to x, we get

Log (1+ x) = c + x – x^2/2 + x^3/3– x^4/4 + x^5/5……………

At x = 0, log 1 = c, or

0 = c, since log 1 = 0. Therefore,

Log (1+ x) = x – x^2/2 + x^3/3 – x^4/4+ x^5/5…………………..(4)

Expansion of log(1+ i tan x)

In the same way, Log (1+ i tan x) can be expanded according to equation (4) as

= i tan x – (-1). Tan^2 x/2 – i tan^3 x/3– tan^4 x/4 + i tan^5 x/5 +…………..

= i ( tan x – tan^3 x/3 + tan^5 x/5 ………..) + ( tan^2 x/2 – tan^4 x/4 + tan^6 x/6 ……) on separating imaginary and real parts.

Proving Euler’s formula by independent approach

Also log (1+ i tan x) = log (cos x + i sin x)/cos x

= log (cos x + i sin x) – log cos x. Therefore,

log (cos x + i sin x) – log cos x = i ( tan x – tan^3 x/3 + tan^5 x/5………..) + ( tan^2 x/2– tan^4 x/4 + tan^6 x/6 ……).

For all real x, cos x is real, therefore, log cos x is also always real.

Comparing real parts,

-log cos x = tan^2 x/2– tan^4 x/4+ tan^6 x/6……

Since cos x is always less than or equal to 1, therefore log cos x is always negative or zero.

Comparing imaginary parts,

Log ( cos x + i sin x) = i(tan x – tan^3 x/3+ tan^5 x/5 ……)………(6)

But tan x – tan^3 x/3+ tan^5 x/5 ………… = x + 2.k.pi, according to equation (3), therefore right hand side of equation (6) equals i.x and equation takes the form,

log ( cos x+i sin x) =i.(x + 2k.pi)

Or cos x + i sin x = e^i.(x+2.k.pi) ………………………..(7).

That is Euler’s formula in general form that has been independently proved.

Interpretation of Euler’s formula

When x is zero, e^i.0 = 1. That is at angle x = 0, it has only real value as 1.

When x is pi/2, e^i.pi/2 = cos pi/2 + i sin pi/2 = i. That means e^i.pi/2 lies on a line perpendicular to horizontal line at a unit length. Perpendicular to horizontal line is Y axis, therefore, e^i.pi/2 denotes a point on Y axis at a unit length. Or i is a point on Y axis at unit length. In other words i is synonymous with Y axis but has a unit magnitude.

From above, it is proved that imaginary part constitutes the magnitude of the line along Y axis and real part along horizontal line.

When x is pi/4, e^i.pi/4 = .707 + .707. i. That is e^i.pi/4 has magnitude .707 on X axis and .707 on Y axis.

When x is pi/2, e^i.pi/2 = cos pi/2 + i sin pi/2 = i. Therefore, e^i.pi/2 is a line coinciding with vertical or Y axis and has unit magnitude.

A generalised statement thus can be made that imaginary part which is unreal, can be plotted along Y axis and is expressible for all mathematical calculations. Also real part is plotted along X axis. Coming to the opening paragraph, I submit that this is due to Euler, why imaginary quantity is expressible on Y axis. But if it is negative, it is plotted on – Y axis and similarly negative real quantity is plotted on – X axis.

Interpretation of e^ix versus e^x

                         Figure showing geometrical representation of e^i.x

Let us see what happens when value of x is increased.

When x is zero, then e^i.0 = 1. When x = pi/2, e^i.pi/2 = 0 +i sin pi/2 = i and it lies along Y axis at unit length. When x = pi, e^i.pi equals – 1 + i.0 = -1 and it is real. It is a point that lies along minus X axis at unit length. Thus as x increases, angle x increases and the point cos x + i sin x moves along a circle of unit length. If x increases from 0 to 2.pi, the point starting from x axis completes full circle and comes back to its starting point.

Let us find out the value of 20^i which can be written as

y = 20^i. Taking log of both sides,

log y = i log 20 or y = e^(i.log 20) = cos (log 20) + i.sin (log 20).

Therefore y has a unit magnitude and denotes a point at an angle of log 20 with horizontal or real part equal to + cos (log 20). It has vertical or imaginary part equal to + sin (log 20) along Y axis.

Similarly y = 40^i, has a unit magnitude, denotes a point at an angle of log 40 with horizontal or real part equal to + cos (log 40). It has vertical or imaginary part equal to + sin (log 40) along y axis.

That means as real base (to the imaginary power) increases, its angle changes whereas magnitude of the point remains unchanged as unity.

                                    Curve of Exponential function y = e^x

But in exponential function, y = e^x, when x increases, magnitude of y increases exponentially and it tends to Infinity when x tends to Infinity. When x is zero, it equals e^0 or 1 and when x tends to minus infinity, it tends to zero. That is e^x increases exponentially from 0 to infinity when x increases from minus infinity to plus infinity.

However in case of e^i.x, when x increases from 0 to infinity, point rotates from angle 0 to 2.pi completing a circle then next circle so on but its magnitude remains unity as we have seen in preceding paragraph. When x changes from 0 to minus infinity, the point e^i.x will circle in reverse direction keeping its length constant as unity.

Locus of point e^ix

We have seen that when x varies from 0 to 2pi, point e^ix rotates from angle 0 to 2pi keeping its length as unity. When x varies from 0 to 4.pi, point completes two rotations in a unit radius circle. Therefore, e^ix represents a circle with unit length when x is variable. If the point is M.e^ix, where M is any real quantity, it corresponds to M.cos x +i. M sin x which has magnitude M. Therefore M.e^i.x rotates in a circles of radius M as x varies.

Simple Harmonic Motion

When point e^i.x rotates in a circle with variation of x from 0 to pi/2, then to pi, then to 3pi/2, then to 2.pi, its projection on X axis changes from 1, then to 0, then to -1, then to 0 and finally to 1 ie in accordance with sin x. This projection on x axis is Simple Harmonic Motion, SHM from its definition that it is the projection of a point (on X or Y axis) that moves in a circle.Similarly, projection of point e^i.x on Y axis is also SHM. Therefore when x varies, point e^i.x moves in a circle and its projection on X axis or Y axis is SHM.

Wave Equation

We have proved that e^i.x equals cos x+ i sin x.

Let y = e^ix = cos x + i sin x or we can say that cos x + i sin x is a root of y. Assuming that function y has real coefficients, it must have another root which is conjugate of cos x + i sin x. Conjugate of cos x + i sin x is cos x – i sin x, therefore, y has another root as cos x – i sin x.

That means if there is a root e^ix of a function with real coefficients, then there must be another root e^-i.x of that function. And solution of that function y is e^i.x + e^-i.x which is equal to 2.cos x.

But y = 2. Cos x is wave motion in a plane. That means if e^ix is a root of a function y, then it must have other root as e-i.x and the function y is a wave equation.


Binomial expansion and derivative of a function are derived ab initio. Based on these primary derivations, Euler’s formula e^i.x equal to cos x + i sin x is proved from expansion of log (1 + i.tan x).

‘i’ which is square root of -1, is synonymous with anti-clockwise rotation by an angle pi/2 or Y axis but it has a unit length. Similarly -‘i’ is synonymous with anticlockwise rotation by an angle 3pi/2 or – Y axis but it also has a unit length. That is positive imaginary quantity is plotted along positive Y axis and negative imaginary quantity is plotted against – Y axis, likewise, positive real quantity on + X axis and negative real quantity on – X axis.

If x is variable, e^i.x denotes a point that revolves along the circumference of a circle with unit radius and represents a unit circle. When x increases, e^i.x represents a point with increased angle x but its magnitude remains constant as unity. When x varies from 0 to 2.pi, point e^ix completes a full circle of unit radius. n^i always has magnitude one, however large or small n is except n is zero or infinity.

Whereas in case of e^x, when x increases, value of e^x increases exponentially. It is zero, at x = Infinity, one at x = 0, Infinity at x = infinity.

Locus of point e^i x is a unit circle when x is variable.

Projection of point e^i.x on X axis or Y axis is SHM when x is variable.

If e^i.x is a root of function y, then e^-i.x must also be a root of y and function y is then a wave equation.

Before parting with this paper, if I do not devote a few lines to the beauty of Euler’s formula and equation, I would be failing in my objective of this submission. For the sake of brevity, I did not explain their application in circuit analysis, electrical equipments and other related fields. I would be mostly correct if I make a sweeping statement that these pervade all fields of sciences. I am sanguine with professor Keith Devine of Stanford University mathematics professor who once said, “like a Shakespearean sonnet that captures the very essence of love, or a painting that brings out the beauty of the human form that is far more than just skin deep, Euler’s equation reaches down into the very depths of existence”


1. Title Image courtesy author Gunther and derivative work Wereon at

2. Image curve of exponential function courtesy author Peter John Acklam at 

3. Courtesy mathematician Keith Devine for his quote at 


Writer is an Electronics and Electrical Communication Engineering graduate and was earlier Scientist, then Instrument Maintenance Engineer, then Civil Servant in Indian Administrative Service (IAS). After retirement, he writes on subjects, Astronomy, Mathematics, Yoga, Humanity etc.


Author: Narinder

I am graduate in Electronics and Electrical Communication Engineering from Punjab Engineering College, Chandigarh. I worked as Scientist in Solid State Physics Lab DELHI then National Fertilizers Ltd , Bhatinda as Instrumentation Engineer, then Ministry of Labour, Employment, Training as Deputy Director. Thereafter, I joined Civil Services in the year 1986, worked in different capacities on administrative posts and retired on September 30, 2013 as Secretary to Government Punjab from Indian Administratve Services. My interests are Astronomy, Physics, History, Music, Law, Spirituality, Administration and writings. I believe in hard work, determination and consistency in efforts. I love to write on topics related to Astronomy Daily life experience and human sufferings. My favourite writers are Leo Tolstoy, Rabindranath Tagore, Mulk Raj Anand and Munshi Prem Chand.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s