Complex Plane
Often cited as the most beautiful equation in mathematics, Euler's Identity is remarkable for its ability to link five of the most fundamental mathematical constants in a single, concise formula. It is a special case of Euler's formula, which establishes the deep relationship between trigonometric functions and the complex exponential function.
To truly understand $e^{i\pi} = -1$, we must visualize its foundation: Euler's formula, $e^{i\theta} = \cos(\theta) + i\sin(\theta)$. The number $e$ represents continuous growth. The imaginary unit $i$ represents a 90-degree rotation. So, $e^{i\theta}$ describes a process where you are constantly "growing" sideways.
Press play and watch the main vector. Notice the small, yellow velocity arrow at its tip. This arrow is always perpendicular to the main vector, forcing it into a circular path. This is the dynamic heart of Euler's formula, and it's why an exponential function can describe a circle.
While the dynamic visualization provides a beautiful intuition, the rigorous proof of Euler's formula comes from calculus, specifically from Taylor series. A Taylor series is a way to represent a function as an infinite sum of its derivatives. By comparing the series for $e^x$, $\cos(x)$, and $\sin(x)$, the relationship becomes undeniable.
First, let's recall the fundamental Taylor series expansions:
For $e^x$:
$$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots $$For $\cos(x)$:
$$ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots $$For $\sin(x)$:
$$ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots $$Now, the crucial step is to substitute $x$ with $i\theta$ in the series for $e^x$:
$$ e^{i\theta} = 1 + (i\theta) + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \frac{(i\theta)^5}{5!} + \cdots $$We simplify this using the powers of $i$ (where $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, etc.):
$$ e^{i\theta} = 1 + i\theta - \frac{\theta^2}{2!} - i\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + i\frac{\theta^5}{5!} + \cdots $$Next, we group the terms with $i$ (the imaginary part) and the terms without $i$ (the real part):
$$ e^{i\theta} = \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots\right) + i\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right) $$Look closely. The first group of terms is the exact Taylor series for $\cos(\theta)$. The second group is the exact Taylor series for $\sin(\theta)$.
Therefore, we have proven that: