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#### Sorunome

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« 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

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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#### Matrefeytontias

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

#### 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.