October 2, 2024 (Updated on: March 7, 2025)

Proof of Euler's Identity $e^{i\pi} +1 = 0$ Using Taylor Series

Euler’s identity is a famous equation in mathematics :

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

It connects five of the most fundamental numbers in mathematics : $e$, $i$, $\pi$, $1$ and $0$.

Additionally, three types of numbers are represented in the identity : integers, irrational numbers and imaginary numbers. This equation is derived from Euler’s formula, which establishes the fundamental relationship between trigonometric functions and the complex exponential function :

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

By setting $x = \pi$, it becomes : $e^{i\pi} = -1$

This equation represents a specific point on the unit circle, defined as a circle with a radius of 1 and centered at the origin $(0,0)$. The angle associated with this point, measured from the positive real axis, is $\pi$ radians (as illustrated in the figure above). $e^{i\pi}$ is a particular location on the unit circle, situated precisely on the negative real axis at an angle of $\pi$ radians.

There are several methods to prove Euler’s identity. The approach based on Taylor series is one of the most used. This method utilizes the Taylor series expansion of transcendental functions, including the exponential function, cosine and sine.

Proof

Taylor series allow a differentiable function to be represented as an infinite sum of terms derived from that function at a given point. For the function $f(x)$ around $x = 0$, the Taylor series expansion is :

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$

where $f^{(n)}(0)$ is the $n$-th derivative of $f(x)$ evaluated at $x = 0$ and $n!$ is the factorial of $n$.

So the Taylor series expansion around 0, valid for all $z \in \mathbb{C}$, of the complex exponential function $e^z$, where $z$ is a complex number is :

$$e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!} = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \cdots$$

This series converges absolutely for all $z \in \mathbb{C}$.

For the complex exponential function, we want to study the case where the argument is imaginary, so, $z = ix$ where $x$ is a real number. We then have :

$$e^{ix} = \sum_{n=0}^{\infty} \frac{(ix)^n}{n!}$$

If we expend the first terms of this series while separating the even and odd powers of $ix$ and using the fact that $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, etc., we obtain :

$$e^{ix} = 1 + ix - \frac{x^2}{2!} - i\frac{x^3}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} - \cdots$$

This series can be rearranged by separating the real and imaginary terms :

$$e^{ix} = \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \right) + i\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \right)$$

The Taylor series expansions of the trigonometric functions $\cos(x)$ and $\sin(x)$, also expanded around 0 and are :

• For cosine :$$\cos(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$$
• For sine :$$\sin(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$

SO, by comparing the series obtained for $e^{ix}$ with the $\cos(x)$ and $\sin(x)$ ones, we have :

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

we can now apply this formula to the case where $x = \pi$. Using the values of the trigonometric functions for $\pi$ :

$$\cos(\pi) = -1 \quad \text{and} \quad \sin(\pi) = 0$$

And we obtain :

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

And finaly :

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