Complex Analysis Complex Analysis Math Forum

 September 7th, 2019, 12:24 PM #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. Last edited by skipjack; October 2nd, 2019 at 03:36 AM. September 7th, 2019, 01:32 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,636 Thanks: 1472 The problem is that $1^i \neq 1$ $1^i = (e^{i2\pi})^i = e^{-2\pi}$ $\left(1^i\right)^{1/4} = e^{-\pi/2}$ Thanks from topsquark September 7th, 2019, 06:41 PM #3 Senior Member   Joined: Oct 2009 Posts: 905 Thanks: 354 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. Thanks from topsquark September 28th, 2019, 09:34 PM   #4
Senior Member

Joined: Aug 2018
From: România

Posts: 110
Thanks: 7

Quote:
 Originally Posted by romsek The problem is that $1^i \neq 1$ $1^i = (e^{i2\pi})^i = e^{-2\pi}$ $\left(1^i\right)^{1/4} = e^{-\pi/2}$
Hello,

I do not understand!From the "WolframAlpha" read:

1) https://www.wolframalpha.com/input/?i=1%5Ei%3D1

2) https://www.wolframalpha.com/input/?...e%5E%28-2pi%29

All the best,

Integrator

Last edited by Integrator; September 28th, 2019 at 09:36 PM. September 28th, 2019, 09:38 PM   #5
Senior Member

Joined: Aug 2018
From: România

Posts: 110
Thanks: 7

Quote:
 Originally Posted by lutgardis 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.
Hello,

https://www.wolframalpha.com/input/?i=i%5Ei.

-----------------------------------------------

1) How much do I do $\displaystyle 0^i$ where $\displaystyle i^2=-1$
2) How much do I do $\displaystyle 0^{i^i}$ where $\displaystyle i^2=-1$

All the best,

Integrator

Last edited by skipjack; October 2nd, 2019 at 03:35 AM. October 2nd, 2019, 01:48 AM #6 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 166 Thanks: 64 Math Focus: Area of Circle $$i^i=e^{- \pi /2}$$ Tags ii11 or 4 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

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