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.

Answer 1

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.

Answer 2

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.

Answer 3

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.