I think there's a more intuitive explanation than the Taylor series expansions (whose correctness, after all, is not exactly obvious). It is not hard to show t

I think there's a more intuitive explanation than the Taylor series expansions (whose correctness, after all, is not exactly obvious). It is not hard to show that the product of two complex numbers in polar form is: (r1, θ1) ⋅ (r2, θ2) = (r1 ⋅ r2, θ1 + θ2) That is, the radii multiply, but the angles add. If we consider only points on the unit circle, where the radii are all 1, clearly the product of any two such points is a third such point. Now consider that e ^ x = lim ((1 + x/N) ^ N) N -> ∞ This equation defines real exponentiation in terms of integer exponentiation, which in turn is defined in terms of multiplication. To make it familiar, consider that (1 + I/N) ^ NP is the formula for compound interest, where I is the interest rate per period, P is the number of periods, and N is the number of compounding intervals per period. Let's say P = 1 so we can ignore P. If the interest rate is 12% annually and you're compounding monthly, then the balance after one year is higher by a factor of (1 + .12/12) ^ 12 ≈ 1.1268249 But you could compound daily, hourly, by the minute, second, millisecond, ... As the compounding interval gets smaller, the result approaches e ^ .12 ≈ 1.1274968 Okay, let's put these two things together: e ^ ix = lim (1 + ix/N) ^ N [let "N -> ∞" be implicit] = lim (r, arcsin x/Nr) ^ N where r = √(1 + (ix/N) ^ 2) As N -> ∞, r -> 1 and arcsin x/N -> x/N, so this reduces to e ^ ix = lim (1, x/N) ^ N = (1, lim (x/N) ⋅ N)) Remember, the radii multiply, but the angles add! This is the key step. From here it's easy: e ^ ix = (1, x) That's the general result; from there the special case is trivial: e ^ iπ = (1, π) = cos π + i sin π = -1 EDIT: clarification.