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)

So exp(-i*Pi) = 1, is this right?

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)

x+iy=o

x=o (or) y=0.therrfore e^0=0

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....

Zero already is a complex number: 0 + 0*i.

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