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When discussing the order relation on $\mathbb{C}$, it is said that such a statement as $z_1 < z_2$ where $z_1, z_2 \in \mathbb{C}$ is meaningless, unless $z_1$ and $z_2$ are real.

My question is, when will a complex number $z$ be real? I know that if $\bar{z}$ is the conjugate of $z$, then

$$z + \bar{z} = 2a$$ $$z\bar{z} = a^2+b^2$$

produce real numbers, but it is easy to add $0i$ to either equation to produce a complex number.

J. Dunivin
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    "strictly real" just means $z=a+0i$ i.e. $z$ happens to lie on the "real axis" of the complex plane. That has no effect on closure of complexes under addition/multiplication. – coffeemath Jun 03 '16 at 21:11

3 Answers3

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A complex is real when its imaginary part is $0$. This is the case iff $z=\bar{z}$

marwalix
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A complex number is strictly real iff $z=\overline{z}$

b00n heT
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    The terminology 'strictly real' appears in your answer and @coffeemath's comment. Why? That is, what is gained by 'strictly real' over just 'real'? (I could understand the useage of 'strictly complex' for a non-real complex number, in which case 'strictly complex' is not equivalent to 'complex'; but this definition of 'strictly real' is equivalent to 'real'.) – LSpice Jun 03 '16 at 21:23
  • because I like to refer to a real number as an element of $\mathbb{R}$... so $z\in\mathbb{C}$ with imaginary part $0$ is not formally a real number – b00n heT Jun 03 '16 at 21:26
  • An element $z \in \mathbb C$ with imaginary part $0$ does lie in $\mathbb R$, and hence is real. I think that distinguishing "an element that is presented as belonging to $\mathbb C$, and that happens to belong to its subset $\mathbb R$" from "an element that is presented as belonging to $\mathbb R$" is not in accord with standard usage. (I want to mumble something about extensionality versus intentionality here, but that may be abusing the terminology.) – LSpice Jun 03 '16 at 21:29
  • http://math.stackexchange.com/questions/389897/the-set-of-real-numbers-is-a-subset-of-the-set-of-complex-numbers

    Is this sufficient? Look at the second answer

     – b00n heT Jun 03 '16 at 21:34
  • To be sure, there are ways of constructing $\mathbb C$ according to which $\mathbb R$ is not literally a subset, but I think that "$\mathbb R$ is not a subset of $\mathbb C$" is also not in accord with standard usage. (See, for example, the comment http://math.stackexchange.com/questions/389897/the-set-of-real-numbers-is-a-subset-of-the-set-of-complex-numbers#comment834395_389904 on the answer http://math.stackexchange.com/a/389904/87579 to which I think you refer.) By the same token, one might say that the real number $1$ is not a natural number, which might not be wrong but is confusing. – LSpice Jun 03 '16 at 21:37
  • Again... I am not saying that you must follow my way; but I just like to emphasize the fact that we are dealing with complex numbers with imaginary part $0$, by using the strictly thingy... – b00n heT Jun 03 '16 at 21:42
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    This is a good point, and I'm sorry if I sounded argumentative. I actually meant to address, and should explicitly have done so, my follow-up mainly to @BenedictVoltaire , to avoid any confusion with others who might use just 'real' in the same circumstance. – LSpice Jun 03 '16 at 21:44
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Complex numbers are real iff $z=\bar{z}$. For $z=a+bi$ we also say that $z$ is real if $b=Im(z)=0$.

Related : Proving complex z is real iff $z=\bar{z}$

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David_Shmij
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