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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? –  Alexander Thumm Sep 16 '11 at 16:32

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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. –  Mulu Sep 16 '11 at 16:49
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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. –  J. M. Sep 16 '11 at 16:57
    
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 $r$). –  Steven Stadnicki Sep 16 '11 at 17:15
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@Steven: There's also the fact that unreal roots of polynomials with real coefficients can only come in conjugates. –  anon Sep 16 '11 at 17:32
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@Steven: I'd say a main reason to consider $\bar{z}$ is that it's a ring homomorphism. –  Qiaochu Yuan Sep 16 '11 at 17:34

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