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I know that $z\in\mathbb{C}/ \mathbb{R}$ means that the domain is the complex plane with the real line removed.

What does the notation $z\in\mathbb{C}\backslash \mathbb{R}$ mean?

EDIT:

Turns out that I got things backwards. If $\mathbb{C}\backslash \mathbb{R}$ is the set difference, what does $\mathbb{C}/ \mathbb{R}$ actually mean?

Kagaratsch
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    You are wrong, $\setminus$ - denotes the set difference operation, while $/$ - denotes the factorization operator. – ald.li Jun 10 '16 at 17:32
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    In fact $\Bbb C\setminus \Bbb R$ means what you think $\Bbb C/\Bbb R$ means. – David C. Ullrich Jun 10 '16 at 17:32
  • @ald.li Could you elaborate on what a "factorization operator" is? – Kagaratsch Jun 10 '16 at 17:34
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    A standard meaning of $\mathbb{C}/\mathbb{R}$ could be the quotient space, when $\mathbb{C}$ is considered as a $\mathbb{R}$ vectorspace for example. – quid Jun 10 '16 at 17:35
  • The forward slash is quotient. Informally this means two things in the numerator space are going to be considered equal when they differ by something entirely in the denominator space. – amcalde Jun 10 '16 at 17:36
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    ${\bf C}/{\bf R}$ is not exactly clear. But I would understand that it is the quotient of ${\bf C}$ by ${\bf R}$ as vector spaces or abelian groups (isomorphic to ${\bf R}$ either way). – tomasz Jun 10 '16 at 17:37
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    @Kagaratsch It seems that your question spawned a lot of beautiful answers and comments. D_S's answer is great, but as tomasz pointed out, the quotient might mean slightly different things depending on the structure you consider on the original spaces. – ald.li Jun 10 '16 at 17:43

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$z \in \mathbb{C} \setminus \mathbb{R}$ means that $z$ is a complex number that is not a real number. I.e., any number of the form $a+bi$ where $b \not= 0$.

The "backslash" $\setminus$ is the set-difference or set-minus operation. In general $A \setminus B$ is the set of all $x \in A$ such that $x \not\in B$.

The "forward slash" $/$ is a quotient operator. $\mathbb{C}/\mathbb{R}$ would be the set of all cosets of $\mathbb{R}$ in $\mathbb{C}$; these cosets each contain complex numbers whose imaginary components are equal.

Brian Tung
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  • So for each real $b$ in $z=a+ib$ there is a space of real $a$'s, and $\mathbb{C}/\mathbb{R}$ is the collection of all these spaces? Then I kind of don't see the difference between $\mathbb{C}/\mathbb{R}$ and $\mathbb{C}$, since both sets seem to contain elements $z=a+ib$ with arbitrary real $a,b$. Could you clarify? – Kagaratsch Jun 10 '16 at 17:43
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    @Kagaratsch: $\mathbb{C}$ is the set of complex numbers—the set of points in the complex plane, so to speak. $\mathbb{C}/\mathbb{R}$ is a set of disjoint subsets of $\mathbb{C}$—the set of horizontal lines in the complex plane, as it were. – Brian Tung Jun 10 '16 at 18:22
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    @Kagaratsch: It's analogous to the difference between the set of squares on a chessboard, and the set of rows on a chessboard (ranks, if you prefer). They both cover the same squares, but squares are only directly elements of the first set. The second set contains things that in turn contain squares. – Brian Tung Jun 10 '16 at 18:26
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    A division slash is used in this context as a sort of mnemonic, since $A / B$ is the set of cosets of $B$ in $A$. It divides $A$ up into copies of $B$. For instance, if $A$ is the set of integers from $0$ to $20$, and $B$ is the set of multiples of $3$ from $0$ to $18$, then $A / B$ is a set of three cosets of $B$: $B$ itself, then ${1, 4, \ldots, 16, 19}$ ($B$ with "1 added", so to speak), then ${2, 5, \ldots, 17, 20}$ ($B$ with "2 added"). The cardinalities work out: $21 / 7 = 3$. But unlike ordinary division, the structure of the sets is preserved, to the extent possible. – Brian Tung Jun 10 '16 at 18:45
  • Thank you, I think now I understand the concept better. – Kagaratsch Jun 13 '16 at 18:23
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Actually, $\setminus$ is the set difference.

Nathan
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$\mathbb{C}/\mathbb{R}$ is the set of subsets of $\mathbb{C}$ of the form $$z + \mathbb{R} = \{z + r : r \in \mathbb{R} \}$$ where $z$ is a given complex number. Two sets $z+ \mathbb{R}$ and $w+ \mathbb{R}$ might be equal, even if $z \neq w$. You can show that two of these subsets $z + \mathbb{R}$ and $w+ \mathbb{R}$ are equal if and only if the difference $z - w$ is a real number, i.e. if $z$ and $w$ lie on the same horizontal line parallel to the real axis.

D_S
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  • But if $\mathbb{C}/\mathbb{R}$ is the set of all such lines parallel to the real axis, then it seems to me that it contains the complete $\mathbb{C}$. What is the difference? – Kagaratsch Jun 10 '16 at 17:48
  • The elements of $\mathbb{C}$ are points, and the elements of $\mathbb{C}/\mathbb{R}$ are lines. That's a big difference. – D_S Jun 10 '16 at 20:42
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It depends on the context (and the course) in which you are working.

As many people have mentioned already, if you are treating $\mathbb{C}$ and $\mathbb{R}$ as vector spaces (in an abstract algebra or linear algebra class, for example), then $\mathbb{C}/\mathbb{R}$ is the set of cosets of $\mathbb{R}$ in $\mathbb{C}$.

If, however you are in the context of topological spaces (as in a topology course), then $\mathbb{C}/\mathbb{R}$ is the space of all complex numbers viewed as points, where the line of real numbers has been collapsed down to a single point.You could draw the space as an hourglass where the middle point of the glass is the collapsed real line. The important difference here is that in topology, saying that all the real numbers are going to be treated as the same point does not cause any of the other complex numbers to be equal to each other. In the algebraic quotient, it does.

in the algebraic quotient we would have $4 + 7i = 9 + 7i$ since they differ by the real number 5. In the topological quotient, these two numbers would stay unequal because they are not real numbers.

Mike Martinez
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