Say I have three numbers, $a,b,c\in\mathbb C$. I know that if $a$ were complex, for $abc$ to be real, $bc=\overline a$. Is it possible for $b,c$ to both be complex, or is it only possible for one to be, the other being a scalar?
[Math] Can the product of three complex numbers ever be real
complex numbersexamples-counterexamplesreal numbers
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To solve polynomials, the answer is no. The Fundamental Theorem of Algebra says that every (non-constant) polynomial, of degree $n$, with complex coefficients has $n$ roots (including repeated roots) in $\mathbb{C}$.
As a result of this, the set of complex numbers, unlike the set of real numbers, is algebraically closed, which means that we cannot 'escape' $\mathbb{C}$ using any elementary operations, like $+, - , \times, \div, \sqrt{}, e^{...}$ etc. So, in this sense, we don't really 'need' to extend the complex numbers.
However, there do exist hypercomplex numbers, like the quaternions, $\mathbb{H}$, which, instead of using just one imaginary unit $i$, use three: $i, j, k$, each satisfying:$$i^2=j^2=k^2=ijk=-1$$
Quaternions are used in modelling 3D vectors, and have a lot of use in 3D mechanics. A useful property of a quaternion is that it does not satisfy commutativity: e.g. $i\times j \neq j \times i$.
N0w, if you still want to go further, you can explore the octonions, $\mathbb{O},$ which are an extension of the quaternions. Octonions have 7 imaginary units: $e_1, e_2, ...e_7$.
Octonions have far fewer (that I know of, at least) practical uses that quaternions. Octonions also have the interesting property that they lack associativity: e.g. $x\times(y\times z) \neq (x \times y) \times z$ (for $x,y,z \in \mathbb{O}$).
And, finally, at least as far as mathematical interest has gone, we have the sedenions, $\mathbb{S}$. These are an extension of the octonions and have 15 imaginary units: $e_1,e_2,e_3,...,e_{15}$. The special property about the sedenions is that they have zero-divisors (meaning that there exist non-zero sedenions $x$ and $y$ such that $xy=0$). Interesting, this property, isn't it? Not particularly, intuitive, to say the least.
Now, I'll leave as an exercise for you to find out about the applications of hypercomplex numbers, and how to multiply them (hint: Google 'Fano plane mnemonic', which explains how to multiply octonions- this same idea can be extended to the sedenions).
You can put (partial) orders on the complex numbers. One choice is to compare the real parts and ignore the complex ones. Another is to use the lexicographic order, comparing the real parts and then comparing the imaginary ones if the real parts are equal. Another is to use the modulus. There are many more. The distinction with the order on the reals (or subsets of the reals) is that the order relation is compatible with addition and multiplication. You can't do that in the complex numbers. The simple proof is to ask whether $i$ is greater or less than $0$. In either case, $i^2=-1$ should be greater than zero.
Best Answer
For example $z^3=1$, where $z\neq1.$
Id est, $$a=b=c=-\frac{1}{2}+\frac{\sqrt3}{2}i.$$