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September 7th, 2019, 11:24 AM   #1
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Now I do know that i^i=e^(-pi/2+(2*n*pi)) and has infinite solutions where n is an integer plus the solutions are all real.
But I recently asked this question to my friend and here is her reply:-
Now 1^(1/4) has four solutions i, -i, 1 and -1 but none of them satisfy the solution of i^i.
I can't find a mistake in her solution. Can somebody please help me out? This problem is bugging me for a while. Forgive me if I did some silly mistake. I am not really bright in maths.
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September 7th, 2019, 12:32 PM   #2
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The problem is that $1^i \neq 1$

$1^i = (e^{i2\pi})^i = e^{-2\pi}$

$\left(1^i\right)^{1/4} = e^{-\pi/2}$
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September 7th, 2019, 05:41 PM   #3
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The law $a^{bc} = (a^b)^c$ is a dangerous one. It holds for real numbers, but not necessarily for complex numbers! Well, perhaps it holds in some form with multivalued functions, but it needs a proof.
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