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?
Best Answer
Since I feel like your question is trying to address finding some kind of generalisation of the fundamental theorem of algebra (which can be rephrased as saying any nonconstant complex polynomial is surjective), I think one of the best things you can get is Picard's little theorem.
First, let me mention that continuous complex functions are not necessarily surjective, even if they aren't constant: for example, the absolute value function $|\cdot|:\Bbb C\to\Bbb R_{\geq0}\subset\Bbb C$ is certainly continuous, but is also certainly not surjective. Therefore, just having continuity is not enough, so if we want to remedy this by making the functions in question look "more like polynomials", then we ought to make them smoother; that is, (complex) differentiable.
It turns out being complex differentiable is quite a bit to ask: unlike in the real case, a function that is complex differentiable will automatically be analytic; that is, it will have a Taylor series expansion at the point where it is differentiable. Therefore, differentiable complex functions can be thought of as "infinite degree polynomials", and we can go back to the question of: does the fundamental theorem of algebra somehow generalise to this setting?
Cutting to the chase, one place we might end up is Picard's little theorem, which says that if our complex function is differentiable everywhere and also not a constant, then its image will be just about surjective; that is, its image will be $\Bbb C$ except possibly a single point. Therefore, given a complex function $f:\Bbb C\to\Bbb C$ that is differentiable, you will be able to solve $f(z)=a$ for all $a\in\Bbb C$ with at most one exception.
For the record, an example of an entire function whose image is missing a point would be the exponential map $\exp:\Bbb C\to\Bbb C$, whose image is $\Bbb C\setminus\{0\}$.