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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?

Truth-seek
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    What I think you're trying to get at here is: is it completely impossible to write out an equation using complex numbers that doesn't have a complex solution? But the thing is, think about what "using complex numbers" means. It certainly doesn't mean "an equation of the form $f(z)=0$ where $f$ maps $\mathbb C$ to $\mathbb C$". You could define any kind of crazy discontinuous function in a way that completely ignores the structure of $\mathbb C$ and just treats it like a set. – Jack M Jul 25 '20 at 08:07
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    ...adding continuity isn't much better, because then you may as well just be dealing with $\mathbb R^2$. What makes $\mathbb C$ $\mathbb C$ are the operations defined on it, especially multiplication. The FTA says that any statement written using only complex numbers, $+, \cdot, -, /$ and $=$ is true for at least one complex number (except when it's logically contradictory, i.e. $1=2$). In other words, any equation built up from the operations has a solution. I think that this is already exactly as general as you want it to be. – Jack M Jul 25 '20 at 08:10
  • Yes, you are right. But, I was thinking if something interesting can be said about continuous or differentiable function (at the very least) like Picard's theorem – Truth-seek Jul 25 '20 at 08:11
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    But can you really say a function is "written using only complex numbers" just because it's complex differentiable? Like I said, you definitely can't just because it's continuous. – Jack M Jul 25 '20 at 08:14
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    @Jack M Maybe not. But, FTA seems inadequate. What if there is a non polynomial equation in complex numbers that needs solving, will we have to extend complex numbers as well. In what sense is complex numbers the end of extension of number systems? Only FTA? – Truth-seek Jul 25 '20 at 12:56
  • But in what sense is an equation "in complex numbers" if it's not polynomial? Polynomial equations cover all equations that can be written with the complex numbers and the arithmetical symbols. And if your equation involves some more advanced operation than those, is it really "in complex numbers"? Picard's theorem is a reasonable generalization, but even there we can question whether taking a limit of a power series is really an "intrinsically complex-number-ey" operation. – Jack M Jul 25 '20 at 13:12
  • You have an extremely limited view of functions. I can define $f(z)$ for complex $z$ as $f(0)=dog, else f(z)=z^2$. This is not onto because there is no $z$ that has $f(z)=0$. Worse, it doesn't stay inside the complex numbers, but it is a fine function. I think you have two different problems that you are trying to combine into one question. I don't understand the upvotes, and have down voted. – Ross Millikan Jul 26 '20 at 02:46
  • The only operations defined on complex numbers is addition and multiplication. But since complex functions that are differentiable everywhere can be uniquely associated with infinite polynomials through taylor series, Picard's theorem can be seen as a possible generalization of the FTA, ryt? – Truth-seek Jul 26 '20 at 10:38
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    $\frac1x=0$ does not have a solution in complex numbers; similarly $x+1=x$ does not either. Arguably they both might in the extended complex numbers $\mathbb C \cup {\infty}$ – Henry Jul 26 '20 at 16:21

4 Answers4

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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 $\Bbb C\setminus\{0\}$.

Zach Goldthorpe
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  • "smoother" - or, how you learn that a mirror is the roughest thing around, in a sense – The_Sympathizer Jul 25 '20 at 18:47
  • It would be interesting if defining some set $\Bbb Y\supset\Bbb C$ and some function $f(y)$ with $y\in\Bbb Y$ is possible in a way that $f(y)$ is defined for all $y\in\Bbb Y$, bijective and $f(y)=|y|$ if $y\in\Bbb C$. This would be a step comparable to the introduction of the complex numbers... – Martin Rosenau Jul 26 '20 at 08:31
  • @MartinRosenau it would be impossible to make such a map bijective simply because $f|_{\Bbb C}=|\cdot|$ is never injective. Set-theoretically, $\Bbb C$ and $\Bbb Y := \Bbb C\times{0,1}$ are in bijection, so you could pick a bijection $p:\Bbb C\to\Bbb Y$ and then take $f:\Bbb Y\to\Bbb Y$ to the map induced by the maps $f(z,0) := (|z|,0)$ and $f(z,1) := p(z)$ to get a surjection that restricts to $|\cdot|$ on $\Bbb C\cong\Bbb C\times{0}$. – Zach Goldthorpe Jul 26 '20 at 15:11
  • @shibai Sorry. "Bijective" is the wrong term; $z^2$ with $z\in\Bbb C$ isn't bijective either. I wanted to say that it would be interesting if there is a function so that all elements $y\in\Bbb Y$ can be represented as $f(y)$. – Martin Rosenau Jul 26 '20 at 16:12
  • @MartinRosenau yea I figured as much. My comment above was a bit cheap, but it does define an example of such a surjection $f:\Bbb Y\to\Bbb Y$ with $f|_{\Bbb C}=|\cdot|$ (albeit not exactly "respecting" the original function it extends). It might be more interesting once $\Bbb Y$ is further required to have some structure that $f$ respects, though :) – Zach Goldthorpe Jul 26 '20 at 16:17
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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.

Rivers McForge
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While it is not completely clear what you mean by "operations within the complex numbers", it almost sounds that you are alluding to the fundamental Theorem of Algebra. This Theorem says that any polynomial in $\mathbb{C}$ factors completely over $\mathbb{C}$. So, all the roots "stay inside the complex numbers". If you are familiar with Field Theory, this is equivalent to their being no algebraic extensions of $\mathbb{C}$.

However, you second thought still does not hold completely in the complex numbers. You cannot just invert something like $z^2$ because it is not a bijective function on $\mathbb{C}$. To define $\sqrt{z}$ you can write this as $z^{1/2} = e^{1/2\log z}$ where $\log z$ is a branch of the logarithm. Remember that the logarithm is not analytic on all of $\mathbb{C}$ and you you will need to take a branch cut.

If such an inverse exists, then it is necessarily continuous and and analytic where the original function is nonzero (albeit, these are true in the non-complex case as well).

Mike
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  • Is there something more general than the Fundamental Theorem of Algebra (but similar in spirit) that can be said about complex numbers? That's what I am trying to ask. – Truth-seek Jul 24 '20 at 21:37
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    @Truth-seek What do you mean by "similar in spirit"? – Mike Jul 24 '20 at 21:38
  • I am not sure myself. Is there some sense in which we will never have to extend complex numbers to solve any complex equation (more general than polynomial equations) – Truth-seek Jul 24 '20 at 21:40
  • @Truth-seek In an abstract algebra sense, when we talk about solutions to polynomials, they are in some sense artificial. Specifically, this can be done by quotienting out an irreducible polynomial in a polynomial ring (called Kronecker's Theorem). For instance, $\mathbb{C} \cong \mathbb{R}[x]/(x^2 - 1)$. Thus, you are seem to be talking about the theory of transcendental functions. Based on the answers here: https://math.stackexchange.com/questions/1339244/is-there-a-theory-of-transcendental-functions there does not seem to be too much that can be said. – Mike Jul 24 '20 at 21:51
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The definition of the inverse function is, if $$ f : X \to Y$$ then, $$f^{-1} : Y \to X \\ f^{-1} \circ f = f \circ f^{-1} = I$$ (Where $I$ is the indentity function) In the second paragraph of your post, you have written that $\sqrt{x}$ is the inverse of $x^2$ but it is not defined for $-1$, but you didn’t notice that if you consider $\sqrt{x}$ as the inverse of $x^2$ then the inputs/domain of $\sqrt{x}$ must be the outputs/range of the $x^2$ and $-1$ is not in the range of $x^2$ and hence cannot be an input of $\sqrt{x}$.

Complex polynomial functions can have inverses, and existence of inverse does imply continuity.

Knight wants Loong back
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  • I didn't mean inverse in this sense. I think what I meant was the function being onto. I will edit the question. – Truth-seek Jul 24 '20 at 22:08