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If I understand correctly, imaginary numbers were invented in order to expand the domain of the square root function into the negative numbers. Curiously though, no such expansion from the complex numbers is necessary to define the square root (or any other root) for all complex numbers. In other words, any root of any complex number will be a complex number and there's no need to define "imaginary imaginary" numbers for radicals everywhere in the complex domain to make sense. Why is that? And are there functions for which such an extension is necessary?

More formally, why is the set of real numbers not closed under the operation of roots but the set of complex numbers is? And are there any functions F they would require an extension from C to a new field of numbers?

ewokx
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AEStudent51345
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    It was shown soon after their invention that in the complex numbers every polynomial has a zero. – coffeemath Jun 24 '22 at 01:51
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    This is a clear duplicate of why is it that Complex Numbers are algebraically closed? – Toby Mak Jun 24 '22 at 01:53
  • The classic way of looking at $\sqrt{i}$ as a root of $x^4 + 1 = 0$ certainly works, but you could also look at De Moivre's theorem to consider what happens when you take the roots of complex numbers. They are like two sides of the same coin. – Toby Mak Jun 24 '22 at 01:58
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    About the "curiously though" factor, complex numbers are not the only ones to be closed under the operation they extend. Consider for example integers as an extension of naturals with subtraction (or opposite), or rationals as an extension of integers with division (or inverse). – dxiv Jun 24 '22 at 02:44
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    As for "to a new field of numbers", there are certainly number systems larger than complex numbers. Whether or not they are fields is another matter. The quaternions are a common number system to cite but they fail to be a field since you have $ij = -ji$ for example so multiplication is not commutative. Rather, they are said to form a "division algebra." These don't necessarily come about because of a specific function being performed on complex numbers though. – JMoravitz Jun 24 '22 at 03:07
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    If you look at the actual history, there seems to have been very little interest in taking square roots of negative numbers until someone discovered a technique for finding real roots of cubic equations that involved complex numbers in the intermediate steps (but not in the final answer). So it was never about square roots, really, but rather it was about polynomial equations. Which is all the more reason to look at the linked question. – David K Jun 24 '22 at 03:08
  • One definitely can give equations that no complex number satisfies, such as $x+1=x$ (satisfied by $\tilde{\infty}$, extends complex numbers to $\widehat {\mathbb {C} }$, $|x|=i$, satisfied by split-complex unity $j$, extends complex numbers to tessarines, or $x^2=0$, satisfied by dual unity $\varepsilon$, extends complex numbers to dual numbers $\mathbb{D}\mathbb{C}$ https://vixra.org/pdf/1508.0420v1.pdf/dualComplex/index.htm. – Anixx Jun 24 '22 at 04:32

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