Euler's Identity

Written: 31/07/2008 Last edited: 29/07/2009 Filed under: Pure Mathematics

You may have heard of the formula $e^{i\pi}+1=0$ , or $e^{i\pi}=-1$ . Intriguing? Or mystifying? Read on...

Identity Crisis - The "What"

Euler's identity is well renowned because it incorporates 3 mysterious constants integral to pure mathematics, as well as 1 & 0, or -1. These are:

e: The "base of all natural logarithms" - the integral part, and the first 20 decimal places, of e are 2.71828 18284 59045 23536... See this Wikipedia article for more information.
π (pi): The ratio of a circle's circumference to its diameter - to 20 decimal places π = 3.14159 26535 89793 23846. See this Wikipedia article for more information.
i: Also called the "imaginary number", i = √-1. There is no 'decimal expansion' for i, hence its name. Its understanding extended traditional mathematics into "complex mathematics". Again, see the Wikipedia article.

Go Figure - The "Why"

Now, Euler's Identity is a "special case" of Euler's Formula (this means that if a certain set of conditions are satisfied, Euler's Formula turns into Euler's Identity). Consequently, we first need to prove Euler's Formula.

Now, this involves differentiation. If you're not happy with that, I strongly suggest you start with some basic background reading on this subject. WikiBooks makes a great refernce and guide for many subjects. Calculus is no exception, so see the WikiBooks Calculus book's section on differentiation.

When you're happy, particularly with regard to differentiation of the basic trigonometric function, you can begin to analyse Euler's Formula:

Euler's Formula

Firstly, define the function f(x), where $f(x) = {\cos(x) + i\sin(x) \over e^{ix}}$ This is 'legal' to differentiate, since it has a value at all points: $e^{ix} \times e^{-ix} = e^0 = 1$ implies that e^ix at no point equals 0, which would cause a division by zero, making the funcion impossible to differentiate.
Now, differentiate this function to obtain f'(x), making use of the quotient rule:
1. Put $p = \cos(x) + i\sin(x)$ to simplify the writing.
2. Then $\begin{align*} \frac{dp}{dx} &= \frac{d}{dx}(\cos(x) + i\sin(x)) \\ &=\frac{d}{dx}\cos(x) + \frac{d}{dx}i\sin(x) \\ &= -\sin(x) + i\cos(x) \qquad \qquad \text{($i$ is constant)} \end{align*}$
3. Therefore: $\begin{align*} f'(x) &= \frac{ e^{ix} \frac{dp}{dx} - ie^{ix}p }{ (e^{ix})^2 } \\ &= \frac{ \frac{dp}{dx} - ip }{ e^{ix} } \\ &= \frac{ -\sin(x) + i\cos(x) - i\cos(x) - i^2\sin(x) }{ e^{ix} } \\ &= \frac{ -\sin(x) + i\cos(x) - i\cos(x) + \sin(x) }{ e^{ix} } \\ &= \frac{ 0 }{ e^{ix} } \\ &= 0 \end{align*}$
If the function differentiates to 0, then it has a rate of change of 0, and so doesn't change. Hence, we can see that $f(x) = f(0)$ for any x.
So $f(x) = f(0) = \frac{\cos(0) + i\sin(0)}{e^{0i}} = \frac{1 + 0}{1} = 1$ .
Now, if $\frac{\cos(x) + i\sin(x)}{e^{ix}} = 1$ , clearly we have: $\cos(x) + i\sin(x) = e^{ix}$ QED