As we know, complex conjugation is the reflection symmetry of a complex with respect to the real axis. Is there a standard terminology for a pair of complex numbers which have identical imaginary parts and opposite real parts? If not, why such kinds of complex numbers are not as important as conjugate complex so that there is even no special math concept for them?
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5Given a complex number $z:=a+bi$ (with real $a$ and $b$), the complex conjugate $\overline{z}:=a-bi$ is helpful because $z\overline{z}=a^2+b^2=|z|^2$. Flipping the sign of the real part simply doesn't offer comparable utility. – Blue Jul 17 '22 at 03:10
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2@Blue let $z^\star=-a+bi$, which gives $z*z^\star=-(a^2+b^2)=-|z|^2$, which looks also useful... – jsxs Jul 17 '22 at 03:25
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2(As noted in @jjagmath's answer) $z^\star$ is simply $-\overline{z}$; and, from there, we get $zz^\star=z(-\overline{z})=-|z|^2$ easily enough. The benefit of defining $z^\star$ may not be worth the notational clutter. That said, in a particular context, such notation may make perfect sense, so feel free. (One might compare this to how $\sec x$ is $1/\cos x$, which might make $\sec x$ (likewise, $\tan x$ and $\csc x$) seem like "notational clutter"; yet, on balance, I believe these "extra" functions earn their place in the trig books.) – Blue Jul 17 '22 at 03:54
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1Closely related: Why do we not name $-a+bi$ in relation to $a+bi$? – MJD Jul 17 '22 at 17:05
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$a+bi \rightarrow -a+bi$ is not a member of the Galois group. $a+bi \rightarrow a-bi$ is. – Acccumulation Jul 17 '22 at 22:30
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"Negative of conjugate" cones too mind. A nonzero complex nu.ber and its negative of conjugate have the property that their product is a negative real number. – Oscar Lanzi Jul 18 '22 at 00:08
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One key point is that the conjugation map plays reasonably well with the algebraic structure: $\overline{x+y}=\overline{x}+\overline{y}$ and $\overline{x\cdot y}=\overline{x}\cdot \overline{y}$. Put another way, conjugation is an automorphism of the complex field. In fact, it's the only "reasonably nice" (e.g. continuous) nontrivial automorphism of $\mathbb{C}$ at all.
On the other hand, the map $$a+bi\mapsto (a+bi)^\star:=-a+bi$$ is not nearly as algebraically nice. It still plays well with addition, but not with multiplication: for example, $$1^\star\cdot1^\star=1\not=-1=(1\cdot 1)^\star.$$
Noah Schweber
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Hi Sir, thank you! I feel your explanation nearly touches the key point... please let me wait for some other answers – jsxs Jul 17 '22 at 03:30
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6@jsxs: To add on to Noah's point, remember that the complex plane is more than a 2d cartesian plane. The algebraic structure distinguishes the real axis from the imaginary axis. For example, multiplication by some $z∈ℂ$ is a spiral transform where the scale factor is $|z|$ and the angle is that of $z$ measured from the positive real axis. In fact, this viewpoint ties in with conjugation; you can easily see that reflection about the real axis commutes with addition and multiplication, whereas reflection about the imaginary axis does not. – user21820 Jul 17 '22 at 12:14
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@jsxs: You're welcome! Somewhat related to that, you may be interested in this post about negation and multiplication on ℝ. Note that we can motivate ℂ using this viewpoint of ℝ as real-world positions on a line, by simply going from scalings of the line to the more general notion of spiral transforms of the plane centred at the origin, that is motivated by asking whether there is some transform $i$ such that $i∘i = -1$. – user21820 Jul 17 '22 at 17:22
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The operation you are describing is simply $-\overline{z}$, so I'd say it would be overcomplicating things to introduce a new notation. The importance of complex conjugate over this other reflexion comes from the fact that $f(z) =\overline{z}$ is a field automorphism. Actually, it's the only non-trivial field automorphism of $\Bbb C$ that fixes $\Bbb R$.
jjagmath
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Your last sentence is wrong: at least assuming the axiom of choice, conjugation is not the only non-trivial field automorphism of $\mathbb{C}$ - in fact, $\mathbb{C}$ has $2^{2^{\aleph_0}}$-many field automorphisms (fix a transcendence basis and consider any permutation of it). However, conjugation is the only nontrivial continuous (or at all "topologically tame") nontrivial automorphism. – Noah Schweber Jul 17 '22 at 03:07
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You're right, sorry. I was thinking in field automorphisms that are the identity on $\Bbb R$ – jjagmath Jul 17 '22 at 03:09
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"stabilizes $\Bbb R$" will do, since $\Bbb R$ has no non-trivial field automorphisms. – Marc van Leeuwen Jul 18 '22 at 04:41
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Just to state your point in a less precise way ;) The defining property of $i$ is that $i^2=-1$, and that is a property shared by $-i$ (because $(-i)^2=-1$), which is why exchanging $-i$ for $i$ is an interesting operation. – Carsten S Jul 18 '22 at 11:30
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For non-zero complex $z = re^{i\theta}, ~\overline{z} = re^{-i\theta}.$
For non-zero complex $z = x + iy = re^{i\theta}$,
$w = -x + iy$ can be similarly expressed as
$re^{i(\pi - \theta)} = re^{-i\theta} \times e^{i\pi} = -\overline{z}.$
Alternatively, $(x - iy) \times -1 = (-x + iy).$
user2661923
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