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Author Topic: Seemingly Complex Numbers Paradox  (Read 5071 times)

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Offline Sorunome

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Seemingly Complex Numbers Paradox
« on: October 02, 2015, 03:13:57 pm »
So, in uni with a few other students we had a seemingly paradox with complex numbers, but I think we solved it, wanted to share it with you guys anyways:
##i = e^{i\frac{\pi}{2}} = e^{\frac{1}{4}(2\pi i)} = (e^{2\pi i})^{\frac{1}{4}} = 1^{\frac{1}{4}} = 1##
On first and second glance this seems to be math-defying as ##i## is clearly not equal to ##1##. Below is the solution to this paradox we came up with.
Spoiler For Solution we came up with:
As ##i## clearly does not equal to ##1## there must be an issue here. We believe the issue to lie within the ##1^{\frac{1}{4}}##. As this is taking a root the answer is not unique, but there are actually multiple answers to ##1^{\frac{1}{4}}##.
Let me re-format that a bit to make it clear:
##
1^{\frac{1}{4}} = x \\
\sqrt[4]{1} = x \\
1 = x^{4}
##

From this point on it is quite obvious, there are multiple solutions to ##x##, that is ##1##, ##-1## and also ##i## and ##-i##.
So we have to say:
##
1^{\frac{1}{4}} = \{-1, 1, -i, i\}
##
So now clearly our real solution, ##i##, is present!
So yeah, just thought it would be fun to share :P

I am nowhere near xedas math skills, please forgive me
« Last Edit: October 02, 2015, 03:26:05 pm by Sorunome »
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Offline Matrefeytontias

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Re: Seemingly Complex Numbers Paradox
« Reply #1 on: October 04, 2015, 06:26:49 am »
Nice, it got me :P
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Offline Legimet

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Re: Seemingly Complex Numbers Paradox
« Reply #2 on: October 12, 2015, 09:32:08 pm »
Actually, the reason it doesn't work is because 1/4 isn't an integer. If a, b are real and c is an integer, then

##e^{c(a+bi)}=e^{ac}\cdot e^{bci}=e^{ac}(\cos cb + i\sin cb)=(e^a)^c(\cos b + i\sin b)^c=(e^{a+bi})^c##

But note that we used De Moivre's law, which only holds for integers.
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