Why isn't $e^{i\pi} = -1$ a contradiction? $e$ to anything should be positive.

$e^x$ is positive when $x$ is real. When $x$ is complex, the rules change, $e^{i\theta}$ doesn't represent "multiplication of $e$ by itself $i\theta$ times." It represents the value of the Taylor series, which can be negative or even non-real, depending on $\theta$.

What is $e$ raised to a complex number in general?

$e^z = e^{a+bi} = e^a (\cos b + i\sin b)$. The real part $a$ controls the magnitude; the imaginary part $b$ controls the angle.

Is there a 3D version of Euler's formula?

Yes, quaternions. Hamilton extended complex numbers to four dimensions in 1843 specifically to do 3D rotations.

Why $\pi$ specifically?

Because $\pi$ is the angle (in radians) of half a turn. $e^{i\theta}$ traces the unit circle; at $\theta = \pi$ you're halfway around, at $-1$.

Plug $z = i\pi/2$ into $e^z = i$ and raise both sides to the $i$ power: $i^i = e^{-\pi/2} \approx 0.2079$. An imaginary number raised to an imaginary power is a real number.

Visual proof · Flagship

Euler's Identity, Built From Scratch

From "what even is $i$" to $e^{i\pi} + 1 = 0$ in twelve unskippable steps. The most-celebrated equation in mathematics, derived without skipping a single line.

Quick answer

$e^{i\pi} + 1 = 0$ is true because raising $e$ to an imaginary power means rotating in the plane, and rotating by exactly $\pi$ radians lands you at $-1$. Adding $1$ gives $0$. The whole derivation, from "what is $i$" to the identity, takes twelve steps and uses only the Taylor series for $e^x$, $\sin x$, and $\cos x$. No skipped algebra, no "just trust me."

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