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A bar or star symbol is used for reflecting in the real axis (i.e. complex conjugate).

Is there a commonly (or not so commonly) used symbol for inverting the sign of the real part (i.e. reflecting in the Im-axis) ?

I'm thinking of a situation in which reflections in the axes are the main concepts. Let's use ~z and z*. In this situation -z wouldn't be a main concept but if needed could be written as -z=~z*. Rather than just make up ~z, I wanted to know if there was already a symbol for this.

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    @J.M.: can you make this a 2 symbol answer?2011-09-16

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$-\bar{z}$ (or $-z^\ast$ if you're of that persuasion) is fine for your needs; no need for new notation.


Edit 9/29/2011:

One paper refers to the operation $-\bar{z}$ as "paraconjugation" and uses $z^\ast$ for it, but since $z^\ast$ is also often used for conjugation proper, I can't recommend the notation in good conscience. (I was hard-pressed to find other papers using the same term, also.)

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    Thanks, but I really am looking for a symbol if it exists. See the edited version of the question.2011-09-16
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    None that I know of. It's not a particularly common operation, and whatever gain you may get from compressing to a single symbol may very well be offset by you having to explain the blasted thing anyway.2011-09-16
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    It might be worth noting some of the reasons why $\bar{z}$ and $-z$ are so much more common. $-z$ is more convenient because it's the only one of the bunch that's _analytic_ (and in particular, it doesn't flip the orientation of the plane), which means that it has much better properties; e.g., it's differentiable as a complex function. So why $\bar{z}$ over the imaginary-axis flip? I'd say it's because generally we're more interested in the real part of a complex number than its imaginary part, and $\bar{z}$ has a closer relation to that (in particular, the fact that $\bar{r}=r$ for real $r$).2011-09-16
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    @Steven: There's also the fact that [unreal roots of polynomials with real coefficients can only come in conjugates](http://en.wikipedia.org/wiki/Complex_conjugate_root_theorem).2011-09-16
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    @Steven: I'd say a main reason to consider $\bar{z}$ is that it's a ring homomorphism.2011-09-16
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    Agreed, and those are much better reasons than mine.2011-09-16