Re: irrational powers in Wolfram Alpha

Op 04/04/2025 om 04:53 schreef sobriquet:

 Hi!
 In this recent youtube video there is an interesting discussion about irrational powers:
 https://www.youtube.com/watch?v=aYuzwNa0_4o
  One of the claims in the video is that
 1^p = e^(i*2*pi*k*p) for integers k
 and that this will also hold in case p is an irrational number.
 Wolfram Alpha will agree that the statement holds if we pick k = 1 and a rational number like 3/7:
 https://www.wolframalpha.com/input?i=1%5E%283%2F7%29+%3D+e%5E%28i+2pi+ %283%2F7%29%29+
 But if you try to do it with an irrational number like sqrt(2), Wolfram Alpha says it's False (again with k=1).
 https://www.wolframalpha.com/input?i=1%5E%28sqrt%282%29%29+ %3D+e%5E%28i+2pi+%28sqrt%282%29%29%29+
 Is there any alternative way to verify the statement with Wolfram Alpha for irrational numbers p?

Actually Wolfram Alpha doesn't agree and it's kind of obvious since
(x^y)^z=x^(yz) only holds for positive numbers (so obviously not for complex numbers).
pi(3/7) just turns out to be interpreted as function application (the prime counting function which evaluates to 0 for 3/7) rather than multiplication.