My Math Forum i^i=1^(1/4)?

 Complex Analysis Complex Analysis Math Forum

 September 7th, 2019, 11:24 AM #1 Newbie   Joined: Sep 2019 From: Avalon Posts: 1 Thanks: 0 i^i=1^(1/4)? 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:- i^i=(i^(4i))^(1/4) =(i^((4)i))^(1/4) =(1^i)^(1/4) =1^(1/4) 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.
 September 7th, 2019, 12:32 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403 The problem is that $1^i \neq 1$ $1^i = (e^{i2\pi})^i = e^{-2\pi}$ $\left(1^i\right)^{1/4} = e^{-\pi/2}$
 September 7th, 2019, 05:41 PM #3 Senior Member   Joined: Oct 2009 Posts: 865 Thanks: 328 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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