# Euler’s Identity

One of the most beautiful mathematical equations ever discovered is Euler’s identity. It simply gave the relationship between two transcendental numbers which is e, the base of natural logarithm, π or pi, the ratio of circumference to the diameter of a circle and i which is an imaginary number. A number that we less expect which is equal to $\sqrt{-1}$

This identity is a special case of Euler’s formula which is shown below.

$e^{ix} = cos(x) +isin(x)$     where x is in radians.

This formula is true to any value of x. Let x = π

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

In basic trigonometry, we know that

$cos(\pi) = -1$  and $sin(\pi) = 0$

By simple substitution:

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

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

$e^{i\pi } +1 = 0$

Now come to think of it, irrational number is raised to the product of an imaginary and another irrational number results to an integer value?

Another proof by De Moivre’s Theorem:

$(cos(x) +isin(x))^n = cos(nx) + isin(nx)$

We know that  $e^{ix} = cos(x) +isin(x)$

By substitution:

$(e^{ix})^n = cos(nx) + isin(nx)$

$e^{inx} = cos(nx) + isin(nx)$

Let n = 1, x= π

$e^{i(1)\pi} = cos((1)(\pi)) + isin((1)(\pi))$

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

$e^{i\pi}= -1$

$e^{i\pi}+1=0$

### 1 Response

1. geraldap says:

Very good! 🙂 thanks