TITLE: Are all complex functions onto?
QUESTION [13 upvotes]: I am not sure whether this question even makes sense. But I was just wondering whether all inverse operations of functions defined in complex numbers will stay inside complex numbers. (i.e. we don't have to extend the complex number system):
$x^2$ is a well-defined function of real numbers alone and yet there is no real number such that $x^2 = -1$, i.e. there is no inverse, for $-1$ (and so we need complex numbers).
Is there a theorem that says that this kind of thing cannot happen with complex numbers? Maybe all continuous complex functions are onto? Or, maybe all taylor series with complex coefficients are onto?

REPLY [7 votes]: As I understand it, your question is raising an analogy with the fact that $x^2 + 1 = 0$ has no solutions over $\Bbb{R}$, and we need to "invent" a new number $i = \sqrt{-1}$ (that is, adjoin an element with the appropriate algebraic properties to the field $\Bbb{R}$). This gives us the field $\Bbb{C}$. So your question is, do we have to keep "inventing" new numbers [adjoining new elements] $j, k, l, ...$ indefinitely to $\Bbb{C}$ to solve ever broader classes of equations?
As far as polynomials go, the answer is "no". $\Bbb{C}$ is algebraically closed: if you give me any polynomial $p(x) \in \Bbb{C}[x]$ with complex coefficients, I can give you a complex number $z$ for which $p(z) = 0$. $\Bbb{C}$ is also closed under field operations: for any two complex numbers $a_1 + b_1i$, $a_2 + b_2i$, their sum, difference, product, and quotient can all be written in the form $a + bi$ as well (so we don't need to keep inventing new symbols $j, k, l, ...$ to handle basic field operations either).
It's even true that complex numbers have logarithms and roots, although these are generally multivalued, and cannot be defined on all of $\Bbb{C}$ because of singularity issues at the origin. I can also raise a complex number to a complex power, and my result still is in $\Bbb{C}$. Basically, once you add $i$ to the field $\Bbb{R}$, your work of adding elements is complete.