Complex equations with no complex solutions?

Question

Are there complex equations that admit no complex solutions, but rather quaternions or hypercomplex solutions, for example, in complete analogy to, say, the equation $x \times x = -1$ when restricted to the real line?

Edit: I am indeed using "equation" in its broader sense, as I am not restricting it to operations involving exclusively complex multiplications, say.

@GitGud: That works for not having a complex solution, but I don't think it helps for the "implies quaternionic/hypercomplex solutions" part (which I rather doubt to be true.) — Semiclassical, Jul 29 '14 at 23:21
so all complex polynomials have solutions; $\mathbb{C}$ is algebraically closed — JC574, Jul 29 '14 at 23:22

Jam · Accepted Answer · 2014-09-13T19:22:20.060

5

Using the definition of the quaternions, $$x^2=y^2=z^2=xyz=-1\quad\quad\quad x,y,z\in\mathbb{C},\quad (x\neq y)\wedge(x\neq z)\wedge(y\neq z)$$

The complex solutions to $x^2=-1$ are $x=i$ and $x=-i$ only. The equation needs a third solution but no more exist so $x$, $y$ and $z$ can't all be complex numbers at the same time. However, by definition, there are quaternion solutions.

edited Sep 13 '14 at 19:22

answered Jul 29 '14 at 23:27

Jam

10,325

@Winther That's what I'm trying to show; that this equation has no complex solutions. – Jam Jul 29 '14 at 23:29
so what we are really looking at here then is a set of 4 simultaneous equations, namely $x \times x = -1$, etc. and $xyz = -1$, correct? – user44212 Jul 29 '14 at 23:35
@user44212 Yeah, essentially. In the same way that $x^2=-1$ defines $i$ (with no real solutions), I'm using the equation that defines $i$, $j$ and $k$ with no complex solutions. I'm not entirely sure that it's valid but it seems to work. – Jam Jul 29 '14 at 23:42
Actually, it must be valid as there are only $2$ complex square roots of $-1$ (namely $i$ and $-i$) so to have a third solution, the equation has to be with quaternions. – Jam Jul 29 '14 at 23:44
1

This is correct without the last condition, if $x^2 = y^2 = z^2 = -1$ then we know $x, y, z = \pm i$ so WLOG $x = y$ then $xyz = -z \neq -1$. So this example is correct (and great). – CameronJWhitehead Jul 29 '14 at 23:45
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@CameronJWhitehead Thank you very much & also thanks for verifying :) – Jam Jul 29 '14 at 23:47

score 2 · Answer 2 · answered Jul 30 '14 at 03:22

Simple example, find a nonzero solution to $x^2=0$. This is easy in the dual numbers $\mathcal{N} = \mathbb{R}\oplus \epsilon \mathbb{R}$ where $\epsilon^2=0$. Indeed, the equation $x^2=0$ is solved by any element of $\epsilon \mathbb{R}$ as $x=\epsilon r$ for $r \in \mathbb{R}$ has $x^2=(\epsilon r)^2 = (\epsilon r)(\epsilon r) = \epsilon^2r^2=0$.

For hyperbolic numbers, you can find zero divisors. For quaternions, you can find infinitely many solutions to a quadratic equation. The distinction between nonstandard number systems and fields is quite noticeable.

score 0 · Answer 3 · answered Jul 29 '14 at 23:24

The notion of "equation" is very broad…

I guess what you are looking for is the theorem telling you that every polynomial equation with complex coefficients always has a complex solution. So you will never need quaternions or whatever for those equations.

Complex equations with no complex solutions?

3 Answers3