[FabHK]

Not really. Your equation has no unknowns, but if you consider the "e" in your equation as the unknown, and solve it (here I use "a" instead as the unknown, and use "e" in its usual meaning):

We want a^(i pi) + 1 = 0. Now,

a^(i pi) = e^(ln(a) i pi) = e^(i ln(a) pi) = cos( ln(a) pi) + i sin( ln(a) pi),

so we want cos( ln(a) pi) = -1, sin( ln(a) pi) = 0,

so ln(a) = 1, 3, 5, so a = e, e^3, e^5, ...

Thus indeed e^(i pi) + 1 = 0.

But also ln(a) = -1, -3, -5 work, so for example for

a = 1/e = 0.3678794412... > 0, we have

a^(i pi) + 1 = 0;

and of course for a = e^-99 = 1.0112214926104485... × 10^-43 etc.

[cperciva]

Oops, I meant the smallest root greater than 1, not the smallest root greater than 0.

(Off-by-one errors, they're not just for programmers!)