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.
