# Algebaric closure - Are polynomials with complex _powers_ also complex?

If I'm understanding correctly, the field of real numbers is not algebraically closed, because you can set up a polynomial with real coefficients that has roots that are not in the reals. (e.g. the roots of ${x^2 + 1 = 0}$ is not in the reals, since the square root of minus one has an imaginary component)

However, if you make a polynomial with complex coefficients, the roots are always in the field of complex numbers. (although for some cases it might also be in the reals)

i.e. If A, B and C are in the field of complex numbers, then the roots of: $$Ax^2 + Bx + C = 0$$ will be also be guaranteed to be complex. So in a sense the buck stops with the complex numbers.

What I want to know is if that also works for polynomials of the form: $$Ax^D + Bx^E + C = 0$$ where A,B,C,D,E are complex. Are the roots still in the complex numbers?

Edit - I also want to apologise for the basic level of this question. I'm pretty sure the answer is yes, but I haven't found it mentioned elsewhere yet, and I don't trust my intuitions enough at this point.

Edit2: Here's as far as I can get by myself. If a simple polynomial is in the form $$Ax^B$$ Then we can express it replacing the complex B with the strictly real E & F: $$Ax^B = Ax^{E+Fi}=Ax^Ex^{Fi}$$ and since $e^{i\theta} = Cos(\theta) + iSin(\theta)$ that makes: $$Ax^{E+Fi}=Ax^E(cos(x\ln{F})+iSin(x\ln{F}))$$ then: $$Ax^{E+Fi}=Ax^Ecos(x\ln{F})+Ax^ESin(x\ln{F})i$$ Which really looks like it'll behave the same way and only have complex roots, but I don't know how to show.

• Your equation makes sense only if you are able to assign a definite value to an expression $z^w$ with $z,w \in \mathbb{C}$. This is trivially possible for all $w \in \mathbb{Z}$. For all other $w$ you can try to write $z^w = e^{\ln(z)w}$, but you have to give an interpretation for the logarithm $\ln(z)$. You want $\ln(z) = u$ with $e^u = z$, but there is no unique solutiuon since $e^u = e^{u + 2k\pi i}$ for all $k \in \mathbb{Z}$. Of course you can make a choice, but doing so you will get in hot water. – Paul Frost Sep 3 '18 at 9:09