Euler's identity is usually stated as
eiπ = −1
Rewriting slightly, we have eiπ + 1 = 0. This succinct result relating the fundamental concepts of e, i, π, 1, and 0, is a special case of the trigonometric identity eiθ = cos θ + i sin θ, where θ = π.
To establish the identity for all values of θ, we use the Maclaurin power series:
ez = 1 + z/1! + z2/2! + z3/3! + z4/4! + ...
It follows that
eiθ = 1 + iθ/1! − θ2/2! − iθ3/3! + θ4/4! + iθ5/5! − θ6/6! − iθ7/7! + ...
eiθ = 1 − θ2/2! + θ4/4! − θ6/6! + ... + i(θ/1! − θ3/3! + θ5/5! − θ7/7! + ...)
eiθ = cos θ + i sin θ
Note: One may also establish the result using differential equations:
Let f(θ) = eiθ and g(θ) = cos θ + i sin θ
Observe that f'(θ) = i f(θ) and that g'(θ) = −sin θ + i cos θ = i2 sin θ + i cos θ = i (cos θ + i sin θ) = i g(θ)
Finally, f(0) = 1 = g(0).