Shixil commented at 2023-07-31 00:29:06 » #2821184

x is clearly equal to 0 or any integer multiple of 2pi.

Two simple ways of finding this result:
1. For any exponential that is equivalent to 1, its exponent must be equivalent to 0. 0 is a trivial solution for this, even considering the imaginary number in the exponent.

2. Using Euler’s identity, e^{ix} = cos(x) + isin(x). Forcing the right hand side of the equation to be 1 means the imaginary part is equal to 0 and the real part is equal to 1, so you have two smaller equations to solve: cos(x) = 1 or sin(x) = 0. With this, every integer multiple of 2pi (2pi, 4pi, 6pi, etc) also are solutions.

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