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?

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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.

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