do you mean x^0 or 0^x?
if you mean the first case, it follows from Euler's formula: exp(i*Pi) = -1 (with i = sqrt(-1)) that x^0 = 1 (for x < infinity) = -exp(i*Pi)
if you mean the second case then: 0^x = 0 (for x < infinity) = 1 + exp(i*Pi)
no, it's exp(+i*Pi) = -1 (this is Euler's formula)
what you have written is: exp(-i*Pi) = 1/exp(i*Pi) = 1/(-1) = -1
this makes sense, because if you draw a circle on the x-i plane (the complex plane), you can see that the angle Pi = 180° = -180° = -Pi = 0 (so, it lies on the x-axis)
Exp(i*x)=cos(x)+i sin(x) (trygonometric form)
and hence, exp(i*pi)=-1 and exp(-i*pi)=-1 since cos(x)=cos(-x).
e^0=1=cos(2n*Pi)+i sin(2n*Pi), where i=sqrt(-1) and n \in the set of integers....
The real number "0" can also be represented as a complex number of the form a+bi. In this case a = 0 and b = 0. Therefore, the number "0" represented as a complex number would have the following expression 0 + 0i. The exponential of 0, this means e0, when 0 is represented by the complex number 0 + 0i would be of the form e 0 + 0i. Applying the exponentials properties we have:
e 0 + 0i = e0.e0i = 1. e0√₋₁ = 1. e0 = 1.1 = 1 ↔ e 0 + 0i = 1
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