Functions without complex roots, but with quaternion roots Many introductions to complex numbers begin with the question "What are the roots of $x^2 + 1 = 0$?" This function does not have real roots, but does have complex roots.
Are there functions which, in a similar vein, do not have complex roots but do have roots in the quaternions?
 A: This is a good question with an important answer.
The answer is no, because $\mathbb{C}$ has a property called algebraic closure. This means that any degree $n$ polynomial in $\mathbb{C}$ has $n$ factors (though some may be repeated) so the polynomial can always be fully factorised into linear terms. In particular, it means every polynomial has a root, and intuitively, there is nothing 'missing' from $\mathbb{C}$. This is a very important property about $\mathbb{C}$, which is what makes it so useful.
The quaternions don't really have the same relationship to $\mathbb{C}$ as $\mathbb{C}$ has to $\mathbb{R}$, as $\mathbb{C}$ is adding to $\mathbb{R}$ things that are 'missing' in a sense, whereas $\mathbb{C}$ doesn't actually need anything added to it, and the quaternions, $\mathbb{H}$, just add extra roots to polynomials which already have roots.
A: perhaps surprisingly, the answer here mostly depends on what you consider to be a function (and there are several choices that are more or less valid).

*

*a mapping of input points to output points. in this case the question isn't especially interesting because you can construct a function that has or doesn't have roots in any region.


*a continuous function defined using only complex coefficients. here there still are examples e.g. xy-yx=1


*analytic functions. this is a broad class of functions which includes all polynomials, logs, and a bunch of other stuff. I'm not sure the answer in this case.
