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Just as $$\{x \in \mathbb{R}: a \leq x \leq b\}$$ can be written in the more-compact form $[a,b],$ is there an analogous notation for $$\{z \in \mathbb{C}:z=x+yi, x \in[a,b], y \in[c,d]\} \quad ?$$

Pictorially, the set of all $z \in \mathbb{C}$ lying in the green area is the set that I'd like to express in a more concise form: enter image description here

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9 Answers 9

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Maybe just define something as $[a,b]+[c,d]i$ ?

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    $\begingroup$ @RokKralj Why not? There exists interval arithmetic, which we can do on real numbers, why not to do it on complex? $\endgroup$
    – Somnium
    Aug 16, 2014 at 14:07
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    $\begingroup$ Because you can do it in the real numbers, it wouldn't be obvious that you mean this to be complex numbers. $\endgroup$
    – Teepeemm
    Aug 16, 2014 at 18:53
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    $\begingroup$ @Teepeemm We choose one real value from $[c,d]$, multiplying it by $i$ we get imaginary number, then adding any real from interval $[a,b]$ we get complex number. Quite obvious, isn't it? $\endgroup$
    – Somnium
    Aug 16, 2014 at 19:40
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    $\begingroup$ @alexqwx it's not a matter of being sensible, they mean different things! The former (probably needing round brackets) makes $z$ a complex number, the latter makes it an interval. $\endgroup$
    – Sean D
    Aug 16, 2014 at 20:09
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    $\begingroup$ I think that cartesian product ($\times$) would make much more sense than $+$. Plus somehow implies that the operator result type is the same as operand type. $\endgroup$
    – Rok Kralj
    Aug 17, 2014 at 7:51
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Perhaps $\{z \in \mathbb{C}: \operatorname{Re}(z) \in [a,b], \; \operatorname{Im}(z) \in [c,d]\}$.

The complex numbers have no inherent order, so unless you invent something like $[[a+ci, b+di]]$ I know no more compact way to write this.

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  • $\begingroup$ @Alizter: thanks for improving the format of my answer. I didn't know how to get Re() and Im() in normal style, and I suppose I was too lazy to look it up. $\endgroup$ Aug 16, 2014 at 19:14
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As far as I know there is no widely recognized standard notation. Define one in your paper/book/essay if you need it often.

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    $\begingroup$ +1. Do not just use a notation and expect the reader to know what it means. $\endgroup$
    – GEdgar
    Aug 16, 2014 at 13:15
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    $\begingroup$ @GEdgar Better one: Do not introduce notation unless you are 111% sure you need it and it makes your paper easier to follow. $\endgroup$
    – yo'
    Aug 17, 2014 at 17:38
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Since $\mathbb C$ is just $\mathbb R^2$ with specific operations, the notation $[a,b]\times[c,d]$ obviously do the trick.

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    $\begingroup$ This is correct in a sense, but likely to be misleading unless explained. Whether $\mathbb{C}$ "is" $\mathbb{R}^2$ depends on your personal view of the foundations, and readers who prefer to define $\mathbb{C}$ in a different way may be confused. $\endgroup$ Aug 17, 2014 at 21:41
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    $\begingroup$ I admit that the "just" in my answer was a bit provocative... $\endgroup$
    – Taladris
    Aug 18, 2014 at 15:42
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Well, Cartesian product could work here, $\{z\in\mathbb{C}:\Re(z)\in[a,b],\Im(z)\in[c,d]\}$ as $[a,b]\times[ic,id]$.

But I've never seen a standardized notation for something like this, except for circular regions, such as $|z+1|<4$.

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    $\begingroup$ Similar to denoting a set by $|z+1|<4$, you could write $a<Re z<b, c<Im z<d$. $\endgroup$ Aug 16, 2014 at 11:54
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    $\begingroup$ I do not think it works. First of all, an element of the set $[a,b]\times [ic,id]$ is a pair $(x,iy)$, which is not usually canonically identified with a complex number. Then, $[ic,id]$ means nothing in my opinion since you can't define an ordering for imaginary numbers; $i[c,d]$ would look better, but still dubious. Overall, I'd say do not use it unless you define it in advance. $\endgroup$ Aug 16, 2014 at 12:34
  • $\begingroup$ Yeah, that was something I was having trouble with, figuring out how to define that the second set is imaginary. $\endgroup$
    – Silynn
    Aug 16, 2014 at 12:39
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During the Complex Analysis course I took we used $$[z,w]:=\{(1-t)z+tw:t\in[0,1]\},\quad z,w\in\mathbb{C},$$ which coincides with normal intervals when $z,w\in\mathbb{R}$. But you should define this yourself, because people won't know what you mean if you don't.

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    $\begingroup$ This is a line segment connecting $z$ and $w$, not the rectangle. $\endgroup$
    – Teepeemm
    Aug 16, 2014 at 18:54
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    $\begingroup$ This is a pretty common notation, at least in analytic number theory. $\endgroup$ Aug 17, 2014 at 5:53
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    $\begingroup$ +1 This just shows that defining a complex interval to be a rectangle doesn't sound like a great idea... $\endgroup$
    – yo'
    Aug 17, 2014 at 17:39
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You could write it like this:

$\{(x + yi) \in \mathbb{C}: x \in [a,b], \; y \in [c,d]\}$.

Alternatively, choose one of these:

$[a,b] \times i[c,d]$

$[a,b] \times [ic,id]$

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  • $\begingroup$ You could even drop the $\in \mathbb{C}$, since the inclusion could be inferred from the fact $x$ and $y$ are reals. $\endgroup$
    – Rok Kralj
    Aug 16, 2014 at 12:56
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A teacher of mine once used an "interval notation" for boxes in $\Bbb R^n$. If $\vec{a} = (a_1, \ldots, a_n)$ and $\vec{b} = (b_1, \cdots, b_n)$, then: $$]\vec{a}, \vec{b}[ = ]a_1, b_1[ \times \cdots \times ]a_n, b_n[ = \prod_{i = 1}^n ]a_i, b_i[$$ Maybe you can adapt it for what you need.

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Maybe not the answer you're looking for, but you can include the picture and then write:

X

Let $X$ be a rectangle in the complex plane, as depicted.

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