There are various formulas for finding the roots of polynomials of different real degrees. The quadratic formula being the best known. I also know there are analogous formulas for polynomials of degree 3 and 4, but is there a formula for "polynomials" of the degree, lets say, $3i$?
(Roots being where a function equals $0$)
When i say "polynomials of degree $3i$", i mean polynomials like for example, $2x^{3i}+5x^{2i}-x^i+1$, allowing imaginary integer exponents. Having complex exponents however, would be impossible to order as a polynomial, since complex numbers aren't ordered (sadly). But allowing imaginary exponents, also comes with restrictions, as one cant have real and imaginary exponents, because then an ordering problem arises, for example, which term to put first, $x^{2i}$ or $x^2$? However, if we just consider imaginary exponents, is there an analogue formula for finding the root(s) of a function of degree $ni$, if not, is it possible to contruct one?

I know such functions can have roots, one example being $x^i-1$, having a real root at $x=1$.

EDIT: The definition of a polynomial is "an expression consisting of variables (also called indeterminates) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.", so therefore, a "polynomial" of an imaginary degree isn't technically a polynomial, but my question still stands.

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    $\begingroup$ Ordinarily a polynomial is defined where any exponent on the variable is a nonnegative integer. For instance a power of $-1$ on the variable is not allowed. So I would say your expressions are not polynomials. $\endgroup$
    – paw88789
    Dec 15 '18 at 18:02
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    $\begingroup$ This is fairly obvious I think: Let $y= x^i$ and your equation becomes $2y^3+ 5y^2- y+ 1= 0$. Solve that y and then solve $x^i= y$. $\endgroup$
    – user247327
    Dec 15 '18 at 18:05
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    $\begingroup$ If you put $y=x^i$ in your example, you can solve conventionally for $y$, but then you need to take the $i^{th}$ root. This operation, and raising a number to the $i^{th}$ power, can be accomplished through the complex logarithm, which is itself a non-trivial concept, and needs to be defined precisely before the idea of $x^i$ makes proper sense. $\endgroup$ Dec 15 '18 at 18:09

Let's go with yes. It's a bit complicated to straighten several things out here though.

What's the problem?

Firstly, we have to address the question of what $x^i$ means. There are a couple choices.

The easiest choice is of course to restrict to $x$ positive and real, in which case we can define $x^i$ as $e^{i\log x}.$

The other choice is to replace $x$ with $e^z$ with $z$ complex and solve for roots of the equation $$\sum_{n} a_n e^{izn},$$ and say that the equation has roots at $0$ based on whether the polynomial has a constant term or not.

Why is this a problem?

Well how do we define $(-1)^i$? Well $(-1)=e^{\pi i}=e^{3\pi i}$, so $(-1)^i$ should be $e^{-\pi}=e^{-3\pi}$, except wait, those two aren't equal. Thus we run into a problem with complex exponents. See the complex logarithm for more details. (Also there are other ways to resolve this problem, but I chose what I think are the two simplest above).

Solving your equation

Your "polynomial" isn't very interesting to solve, since if the polynomial is $$\sum_{n} a_n x^{in},$$ we can just solve the normal polynomial $$\sum_n a_n y^n$$ and then solve the equations $x^i = y$.

A much more interesting problem (probably, I don't know)

Now you mentioned that you didn't think it made sense to define polynomials with complex exponents, since you couldn't order the exponents. While it's true that there's no natural order for the exponents, there's no reason that they don't make sense. Perhaps it's best to restrict to Gaussian integers though to at least preserve some semblance of sanity.

We could define the "degree" of such a polynomial to be the largest norm among the exponents in the monomials, and ask if it's possible to find the solution to equations with "degree" at most 2.

For example, can we solve $$x^2 + x^{1+i} + x^{2i}?$$

I don't know, I haven't thought about it much, I just wanted to point out that there's no reason you can't ask the question, and that it might be more interesting than the question you did ask.

One approach to start trying to solve such an equation is to let $y=x^i$ again, so we get a two variable equation $$x^2+xy+y^2,$$ defining a curve, and then find the points on this curve that satisfy the constraint that $x^i=y$ for some appropriate solution to the problem of raising numbers to complex powers.

  • $\begingroup$ Finding the roots by letting $y=x^i$, is probably the easiest and simplest way to solve it, and is something im a bit disappointed in myself for not realizing beforehand. Concerning what you said in your last part: "Now you mentioned that you didn't think it made sense to define polynomials with complex exponents, since you couldn't order the exponents", I have to specify, I never said that they didn't make sense (they do), i just said that it was impossible to order them. And it is, because as stated before, complex numbers aren't ordered. $\endgroup$
    – user613706
    Dec 15 '18 at 19:04
  • $\begingroup$ @NilsPhillipTalgö, I was referring to where you said "But allowing imaginary exponents, also comes with restrictions, as one can't have real and imaginary exponents, because then an ordering problem arises, for example, which term to put first, $x^{2i}$ or $x^2$?" I was just trying to get at the idea that the ordering problem you're referring to is illusory, since there's not really a need to order the monomials. Though it is helpful for computations, see Gröbner bases for more on the utility of monomial orderings. $\endgroup$
    – jgon
    Dec 15 '18 at 19:09
  • $\begingroup$ Note, however, that the monomial ordering doesn't need to be natural, it's just useful to pick an arbitrary one (subject to some compatibility requirements). $\endgroup$
    – jgon
    Dec 15 '18 at 19:09
  • $\begingroup$ I see... , and I have to admit, I wasn't specific enough in my wording, what i really ment, was that one can have real and imaginary exponents, but i was thinking of a scenario where you would have a polynomial with some terms having real exponents, and some terms having imaginary ones, then i stated that you can't order them. Well, as you said, you could just choose to order real ones before imaginary ones or vice versa, so that isn't really a problem. But that wasn't my question, what i wanted to know, was how to find the roots of said polynomial as a function, not how to order them. $\endgroup$
    – user613706
    Dec 15 '18 at 19:19

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