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Is there such an operation? What kind of number would the answer be?

There is an operation between real numbers whose answer is not real: $\sqrt{-1} = (-1)^{1/2} = i$. However, operations between complex numbers such as $\sqrt{i} = i^{1/2} = e^{i\pi/4}$ and $i^i = e^{-\pi/2}$ (principal value) result in complex numbers.

It is known that the solutions of algebraic equations with complex coefficients are always complex numbers.

The values of functions such as $i! = \Gamma(1 + i) \approx 0.4980 - 0.1549i$ are also complex numbers. Is there any function whose value is not complex for a complex input?

One thing that came to my mind is $^i i = \underbrace {i^{i^{\cdot^{\cdot^i}}}}_{i \text{ times}}$, the "$i$th" tetration of $i$, but according to this one, although I don't understand it, the value still seems to be complex.

Edit:

I found a duplicate question. The answer there seems to be "there is no such algebraic operation by the fundamental theorem of algebra."

My question is not limited to algebraic operations, but I can't seem to find any such mathematical function. Then the question arises as to why there are none.

I wonder if this is because when we extend numbers from complex numbers to, say, quaternions, we lose properties such as commutativity, so mapping to such a larger domain is not very useful.

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    Welcome. You're essentially asking for a function $f:\Bbb C\to X$ for some $X$ which is "not $\Bbb C$". Such functions are legion. That's also not a satisfying answer, I'm sure. – FShrike Oct 09 '22 at 17:17
  • Yes, I was aware of that concern. I wanted to ask if there are any "standard" operations or functions. And only defined ones, excluding $0/0$, etc. –  Oct 09 '22 at 17:20
  • Do you count real numbers as "not complex"? If so, take $|z|$; if not, take $zj$ with $j\in\Bbb H$, which maps $x+yi$ with $x,,y\in\Bbb R$ to $xj+yk$, if we identify $i\in\Bbb C$ with $i\in\Bbb H$. – J.G. Oct 09 '22 at 17:45
  • @J.G. Real numbers are complex. Is it an operation "between" complex numbers? And is it not "artificial"? For quaternions I found a similar question but no answer. My question wasn't specifically with quaternions in mind. –  Oct 09 '22 at 18:05
  • Thanks for answering my question. I guess what you're hoping for is a map outside of $\Bbb C$ whose definition doesn't mention anything outside of $\Bbb C$, because $zj$ is "artificial" insofar as it has an explicit $j$ factor. – J.G. Oct 09 '22 at 18:08
  • Example : if $z$ is any complex number of form $(x + iy) ~: ~x,y \in \Bbb{R}$, let $~\overline{z}~$ denote the complex conjugate of $z$. That is, $~\overline{z}~ = (x - iy).~$ Then, both of the following expressions are guaranteed to produce a Real number: $$[z + \overline{z}] ~\text{and}~ [z \times \overline{z}].$$ – user2661923 Oct 09 '22 at 20:30
  • @user2661923 I consider real numbers to be complex, i.e., I'm looking for an operation whose the answer is neither complex nor real (neither rational nor integer.) –  Oct 10 '22 at 20:18
  • Index the letters of the alphabet A,B,C,...,Z as $L_0, L_1, \cdots, L_{25}.$ Construct the function $$f:\Bbb{C} \to {A,B,\cdots,Z}$$ as follows: For $$k \in {0,1,2,\cdots,24},$$ if $~k \leq |z| < (k+1),~$ then $~f(z) = L_k.~$ If $~25 \leq |z|,~$ then $~f(z) = L_{25}.$ – user2661923 Oct 11 '22 at 01:27

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When we define a function on a set (the domain), we are essentially free to do so in whichever way we want. We have carte blanche over both the set being mapped to (the codomain) and the function value that corresponds to each argument, with the only restriction (and a very undemanding one at that) being that each argument can map to at most one value.

We can trivially define a function on the complex numbers that is equal to a constant, equal to some set of real numbers, or really anything else.

Jam
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  • Yes, I know that. I wanted to ask if there are any meaningful, "standard" operations or functions that have some name or symbol attached to them. –  Oct 09 '22 at 17:42
  • @13870445 In that case, you will need to sensibly specify what you mean by a "standard" function. Given any function $f$ on the complex numbers, we have that its real part $\operatorname{Re} f(z)$ is purely real. By almost anyone's definition, the real part function is "standard". So, essentially all functions fit the bill. – Jam Oct 09 '22 at 17:48
  • It's certainly vague. I guess it could be "a special function, something of mathematical interest, with a name or symbol attached to it." I consider real numbers to be complex. –  Oct 09 '22 at 18:10