Is e(-iπ) + 1 = 0 the same as e(iπ) + 1 = 0?

[mnada]

For Euler formula :
e^(iπ) + 1 =0
is it the same if we say e^(-iπ)+1 = 0 or not ? (minus sign is included in the exponent)

 

[RUber]

For Euler formula :
e^(iπ) + 1 =0
is it the same if we say e^(-iπ)+1 = 0 or not ? (minus sign is included in the exponent)

It isn't the same, but true nonetheless.

 

[mathman]

It's a special case using $e^{2n\pi i}=1$ for all integers n.

 

[Ssnow]

Another way as the special case of $e^{i\pi +k2\pi i}=-1$ for $k=0$, in general it is true for $k\in\mathbb{Z}$.