# Will this punch a hole in the field of complex number? [closed]

According to this, complex number is algebraically closed, i.e. every polynomial has complex root. What if we allow other type of equations?

I ask this question because equations seemingly can extent number sustem. (From $x+2=1$, we go from natural number to integer, $2x=1$ we have rationals, $x^2+1=0$ we have complex number). Seems that tetration will be the following path, I want to know if there are equation involving tetration (e.g. $x^{x^x+1}-x^x+x^3=1$)that we can create within the complex domain, has no complex root.

• Polynomials have by definition integer exponents, so the "polynomial" you stated is not actually one. (BTW it does have solutions, see here: wolframalpha.com/input/…) – Anonymous Pi Aug 18 '15 at 16:22
• Your thing isn't even a polynomial; polynomials have nonnegative integer exponents. In any case, $e^x$ has no root. – Akiva Weinberger Aug 18 '15 at 16:22
• If you want a bigger field then $\mathbb C$, by the way, you can look up the quarternions. They're of the form $a+bi+cj+dk$, with $i^2=j^2=k^2=ijk=-1$. They're nonassociative; $ij=-k$ but $ji=k$. – Akiva Weinberger Aug 18 '15 at 16:33
• There seem to be several implicit (and incorrect) assumptions behind the sequence of statements in the question, particularly some basic misunderstandings about the definition of a "polynomial" and the implicit assertion that if an equation has no roots, then "the roots must exist in some larger field". Could you please try to clarify what you're asking? – Andrew D. Hwang Aug 18 '15 at 16:33
• Polynomials are particularly well-behaved; they generally work like integers. "Irreducible polynomials," which can't be factored, are the analogs of primes (i.e. $x^2+1$ over $\mathbb R$, $x+1$ over $\mathbb C$ — or any field, actually), Euclid's algorithm works, you have unique factorization, there are always GCDs and LCMs, etc. This ends up being the main reason that you can always extend a field with polynomial solutions. – Akiva Weinberger Aug 18 '15 at 16:45

Those are two separate questions. $e^u=0$ has so complex solution, but there is no apparent way to attach a new solution and extend the complex numbers by it, while retaining useful properties like $e^u e^{-u} = 1$.
If you did add such a $u$, so as to "escape" to something even larger than the complex numbers, what would $u^3$ or $\sin(u)$ be? Are they related to each other in any way?