3 ms·
The most elementary explanation I ever got for why e^(i pi) = -1 is from the definition e = lim_{n -> infinity} (1 + 1/n)^n First, we can generalize this to
by obastani 10y ago
The most elementary explanation I ever got for why e^(i pi) = -1 is from the definition
e = lim_{n -> infinity} (1 + 1/n)^n
First, we can generalize this to
e^z = lim_{n -> infinity} (1 + 1/n)^(z n) = lim_{m -> infinity} (1 + z/m)^m
using the substitution m = zn. Therefore,
e^(i pi) = lim_{m -> infinity} (1 + i pi/m)^m
Now, converting the complex number (1 + i pi/m) in terms of polar coordinates (r, theta) yields
r = (1 + pi^2/m^2) ~ 1
theta = sin^{-1}(pi/m) ~ pi/m
Since the product of two complex numbers is
(r, theta) (r', theta') = (r r', theta + theta'),
we have
(1 + i pi/m)^m ~ (1^m, m * pi/m) = (1, pi) = -1.