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September 7th, 2019, 12:24 PM   #1
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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.
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September 7th, 2019, 01:32 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, 06:41 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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September 28th, 2019, 09:34 PM   #4
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Quote:
Originally Posted by romsek View Post
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.
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September 28th, 2019, 09:38 PM   #5
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Quote:
Originally Posted by lutgardis View Post
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,

From the "WolframAlpha" read:

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.
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October 2nd, 2019, 01:48 AM   #6
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Math Focus: Area of Circle
$$i^i=e^{- \pi /2}$$
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