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Given a complex number $a+bi$, it has a complex conjugate $a-bi$. The product of this complex number with its complex conjugate gives $(a+bi)(a-bi)=a^2+b^2$.

One might imagine flipping the sign of the real part instead of the imaginary part to get a sort of "anticonjugate", resulting in a similar product of $(a+bi)(-a+bi)=-(a^2+b^2)$.

Clearly this "anticonjugate" is the negative of the conjugate. I suspect I've never seen this concept before because of either (1) no finds this anticonjugate useful or (2) the negative of the complex conjugate is considered without giving it a special name.

Is there a different reason why we don't seem to use or consider "anticonjugates"? Or does the above account for this perception?

user21820
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Galen
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    Yes, as you say, your "anti-conjugate" is just minus the conjugate, I don't think there is any need to introduce a separate concept for that. – Captain Lama Jan 10 '22 at 07:01
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    Thank you for providing the background and motivation for your question and showing a possible answer to it. – jimjim Jan 10 '22 at 07:05

2 Answers2

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Complex conjugates are important because $i$ and $-i$ are fundamentally indistinguishable by definition; $i$ is defined to be a number satisfying the equation $i^2 = -1$, but of course $-i$ must satisfy the same equation. So "any fact" which can be stated about complex numbers must remain true if we swap all occurrences of $i$ with $-i$ (though one must take care with "hidden" occurrences). Complex conjugation is therefore a mapping of complex numbers which preserves many algebraic properties.

In contrast, complex anticonjugation as defined in your question does not preserve any useful properties, because $1$ and $-1$ are not fundamentally indistinguishable; $-1$ is not a successor of the number $0$, it is not a multiplicative identity such that $1x \equiv x$, nor does it satisfy any other reasonable definition of the number $1$.

kaya3
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    More specifically and succinctly, complex conjugation is the [sic] non-trivial field automorphism of the complex numbers. (+1) – Andrew D. Hwang Jan 10 '22 at 17:40
  • This in my opinion, is the right answer; the other reply completely fails to explain why the conjugate is important and the anticonjugate isn't. – MJD Jan 11 '22 at 02:06
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    @MJD This one here is a very good answer, which I had +1'd already. That said, I don't think my "other reply completely fails to explain" - at least not what I had set out to explain. There is a historical perspective to this, too, and complex conjugation first emerged for the purpose of solving cubic equations with real coefficients long before abstract algebra even existed. By the time automorphisms of $\mathbb C$ came to be studied, conjugation was proved to be (the only) one, but it was not brought around because of that. – dxiv Jan 11 '22 at 03:34
  • @dxiv Complex conjugates appear in pairs as roots because if one is a root of a real polynomial then the other must be too - so it's pretty much the same reason, just described at a different level of abstraction. – kaya3 Jan 11 '22 at 08:02
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    @kaya3 That's correct, of course. However, the question here was not why conjugates are used and useful, but rather why "anti-conjugates" are not. The historical reason is that "anti-conjugates" would have been redundant, because there is only one such transformation needed to describe all complex manipulations, and that place was taken by conjugates from the very beginning. The original reason why conjugates were preferred was merely convenience related to solving the casus irreducibilis of cubics in the XVI$^{th}$ century. Deeper reasons were found later, but that wasn't my point. – dxiv Jan 11 '22 at 08:14
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There is a tradition in math to avoid (or, at least, minimize) redundancies.

In this case, complex conjugation $\,\overline{a+ib}=a-ib\,$ has been known and used for a long time. It has many applications, from polynomial equations to calculus, abstract algebra, geometry etc.

In contrast, the proposed "anti-conjugate", say we write it as $\,\widetilde{a+ib}=-a+ib\,$, would be a new concept, without any obvious advantage - conceptual or practical. Moreover, it can be easily expressed in terms of the conjugate as $\,\widetilde{a+ib}=-\,\overline{a+ib}=\overline{i\cdot\overline{i \cdot(a+ib)}}\,$. Thus, redundant.

[ EDIT ] $\;$ Side by side summary.

$$ \begin{matrix} & & & \small{\text{conjugate}}\;\overline{\,z\,} & & & \small{\text{anti-conjugate}}\; \widetilde{\,z\,} \\ \small{\text{symmetry}} & & & \small{\text{over real axis}} & & & \small{\text{over imaginary axis}} \\ \small{\text{involution}} & & \small{\text{yes:}} & \overline{\overline{\,z\,}} = z & & \small{\text{yes:}} & \widetilde{\widetilde{\,z\,}} = z \\ \small{\text{distrib over +}} & & \small{\text{yes:}} & \overline{\,z_1+z_2\,} = \overline{\,z_1\,}+\overline{\,z_2\,} & & \small{\text{yes:}} & \widetilde{\,z_1+z_2\,} = \widetilde{\,z_1\,}+\overline{\,z_2\,} \\ \small{\text{distrib over }\times} & & \small{\text{yes:}} & \overline{\,z_1\,\cdot\,z_2\,} = \overline{\,z_1\,}\,\cdot\,\overline{\,z_2\,} & & \color{red}{\small{\text{no:}}} & \widetilde{\,z_1\,\cdot\,z_2\,} \ne \widetilde{\,z_1\,}\,\cdot\,\widetilde{\,z_2\,} \end{matrix} $$

dxiv
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    I feel that it is quite a stretch to call the last expression "easily expressed" :) – lisyarus Jan 10 '22 at 08:44
  • @lisyarus Granted ;-) though it could be shortened to $,\overline{;z;} = f(z),$, $,\widetilde{;z;}=f(i,f(i,z)),$. – dxiv Jan 10 '22 at 08:51
  • If you're multiplying by $i$, you may as well multiply by $-1$ (which is only $i^2$ anyway), without calling $f$ twice. – J.G. Jan 10 '22 at 09:33
  • @J.G. Right, a central symmetry can be written as a composition of two axial symmetries. Just thought I'd list that one for completeness. – dxiv Jan 10 '22 at 09:44
  • @lisyarus, 'easily expressed' is subjective. The expression you're referring to looked straightforward to me at first glance – Jojo Jan 10 '22 at 15:30
  • @Joe Surely it is subjective, that's why my wording is "I feel that" and not something stronger. Good for you, but to me repeated conjugation interwoven with multiplications looks convoluted. – lisyarus Jan 11 '22 at 04:36
  • Wish the downvoters had left a comment why. – dxiv Jan 11 '22 at 09:21
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    @lisyarus: It seems that it is not easy because it is wrong! Lol! c(i⋅c(i⋅(a+b·i))) = c(i·c(a·i−b)) = c(i·(−a·i−b)) = c(a−b·i) = a+b·i and we are back to the start!! In fact, we don't have to do any computation to know that it is wrong, because no sequence comprising rotations and an even number of reflections can ever be equivalent to a single reflection. – user21820 Jan 11 '22 at 18:41
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    @user21820 Oh, indeed! A nice geometric argument, by the way. – lisyarus Jan 12 '22 at 09:40