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

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.