We know that the set $\left\{a+ib:a,b \in \mathbb{Q} \right\}$ is dense in $\mathbb{C}$. Could one give other examples of dense sets in the complex plane?
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5Add or remove any point from your set. Get infinitely many answers this way. – Jonas Meyer Oct 29 '14 at 05:09
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1$\mathbb{C}\setminus\mathbb{R}$ is dense in $\mathbb{C}$. – J. David Taylor Nov 11 '14 at 19:07
7 Answers
To visulaize, your example is as $\mathbb{Q} \times \mathbb{Q}$ in $\mathbb{R}^2$. So you can rotate, move or some nice scales on this set to produce new dense sets! For example $\mathbb{Q} \times (\pi\mathbb{Q})$ or $(\mathbb{Q} + e) \times (\mathbb{Q} + e)$,...

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Here are some examples in which the dense set is the graph of a function from $\mathbb R$ to ${\mathbb R}.$
The graph of any discontinuous function $f:{\mathbb R} \rightarrow {\mathbb R}$ having the property that, for all real numbers $a$ and $b$ we have $f(a+b) = f(a) + f(b),$ is dense in the plane. Such functions are very exotic, however, not even being Lebesgue measurable (let alone Borel measurable). An example of a Baire $2$ function having a dense graph is $f(x) = \sum_{n=1}^{\infty}\left(\frac{1}{2^n}\cdot f_{n}(x)\right),$ where $f_{n}(x) = \frac{1}{x \, - \, r_n}\cdot \sin \left(\frac{1}{x \, - \, r_n}\right)$ (define $f_{n}(r_n)$ to be $0)$ with $\{r_{1}, \, r_{2}, \, r_{3}, \, \ldots \}$ being any fixed enumeration of the set of rational numbers ${\mathbb Q}.$ Note that no Baire $1$ function can have a dense graph, since any function having a dense graph must be discontinuous at every point (and more than this) and every Baire $1$ function has a dense set of continuity points.
(ADDED 13 DAYS LATER) As Yotas Trejos points out in a comment, it is not clear why the example I gave works. Indeed, the example I gave may not even have a dense graph, or worse still, it may not even define a function from the reals to the reals.
A Baire two function with a dense graph is
$$f(x) \;\; = \;\; \limsup_{n \rightarrow \infty} \;\frac{x_1 + x_2 + \cdots + x_n}{\sqrt {n}}$$
where $x_{1},$ $x_{2},$ $x_{3},$ $\ldots$ is the sequence of digits to the right of the decimal point in the nonterminating decimal expansion of the fractional part of $x$ (define $f(x) = 0$ when $x$ is an integer). This is a trivial variation of the Cesàro-Vietoris function in [1] (p. 333), [2], and [4] (pp. 383-384). The graph of $f$ is dense in ${\mathbb R}^{2}.$ In fact, $f$ has the stronger property that for each $x \in {\mathbb R},$ the set of left subsequence limits of $f$ at $x$ is equal to $\mathbb R$ and the set of right subsequence limits of $f$ at $x$ is equal to ${\mathbb R}.$ Note that to have a dense graph it suffices that, at each point $x$ in some set that is dense in ${\mathbb R},$ the set of subsequence limits of $f$ (not necessarily restricted to be from the right side of $x$ or from the left side of $x)$ is equal to ${\mathbb R}.$ (We could replace the last few words with "is equal to a set that is dense in ${\mathbb R}$", but this additional generality is illusionary since the set of subsequence limits at a point is a closed set.)
Incidentally, Beer [1] and Ceder [3] prove related results that are quite interesting and probably little known. Beer (Theorem 3, p. 329) proves that given any function $G:X \rightarrow {\mathbb R},$ where $X$ is a separable metric space, there exists a Baire two function $g:X \rightarrow {\mathbb R}$ such that the (topological) closure of the graph of the function $g$ equals the closure of the graph of the function $G.$ Thus, the existence of a Baire two function $g: {\mathbb R} \rightarrow {\mathbb R}$ with a dense graph follows from the fact that there exists some function (e.g. a non-measurable additive function) that has a dense graph.
Unknown to Beer, Ceder [3] had already obtained this result as a corollary of an even stronger result. Ceder (Theorem 2, p. 4) proves that given any function $G: {\mathbb R} \rightarrow {\mathbb R},$ there exists a Baire two function $g: {\mathbb R} \rightarrow {\mathbb R}$ such that, at each $x \in {\mathbb R},$ the set of left subsequence limits of $g$ at $x$ union $\{g(x)\}$ is equal to the set of left subsequence limits of $G$ at $x$ union $\{G(x)\}$ AND the set of right subsequence limits of $g$ at $x$ union $\{g(x)\}$ is equal to the set of right subsequence limits of $G$ at $x$ union $\{G(x)\}.$ Ceder then states Beer's result (for the case $X = {\mathbb R})$ as a corollary on p. 5.
Brown [2] cites a 1921 paper by Leopold Vietoris, Stetige Mengen, in which the graph of the Cesàro-Vietoris function is shown to be a connected subset of ${\mathbb R}^{2}.$ This paper is the published version of Vietoris' 1920 Ph.D. work (completed in 1919, I think) and it is very important in the history of topology, but Vietoris might (now) be even better known for how long he lived. He was born in 1891 and he obtained his Ph.D. in 1920 (this in spite of his participation in WW I and then becoming a prisoner of war), and yet he died only a few years ago in 2002.
[1] Gerald Alan Beer, The approximation of real functions in the Hausdorff metric, Houston Journal of Mathematics 10 #3 (1984), 325-338.
[2] Jack Bethel Brown, Almost continuity of the Cesàro-Vietoris function, Proceedings of the American Mathematical Society 49 #1 (May 1975), 185-188.
[3] Jack Gary Ceder, The cluster set structure of real functions, Periodica Mathematica Hungarica 12 #1 (1981), 1-10.
[4] Bronislaw Knaster and Kazimierz [Casimir] Kuratowski, Sur quelques propriétés topologiques des fonctions dérivées [On some topological properties of derivative functions], Rendiconti del Circolo Matematico di Palermo 49 (1925), 382-386.

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1@Yotas Trejos: Your comment suggested some other issues to me that need to be worked out, such as whether the function even has the weaker property of being unbounded in every interval. I'll try to think about this in the next few days (might not be until this weekend because of "day job issues" this week) and fix my answer. – Dave L. Renfro Nov 05 '14 at 14:51
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I have asked this in the link
http://math.stackexchange.com/questions/1007224/why-the-following-function-converges
This topic is very interesting for me.
– EQJ Nov 06 '14 at 01:41 -
1@Yotas Trejos: I just posted a "fix" to my answer. I did manage to locate the necessary information this past weekend, but I was too busy to post it yesterday (Monday). – Dave L. Renfro Nov 11 '14 at 18:57
If $q$ is a positive irrational number, then $A=\left\{nq-k:n,k\in \mathbb{N}\right\}$ is dense in $\mathbb{R}$; therefore, the set $\left\{a+ib:a,b\in A\right\}$ is dense in $\mathbb{C}$. What I mean in my question is to find some unpopular examples like this one.

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The image of $\exp(1/z)$ around any punctured disk centered at zero. (Cf. the Casoratti-Weierstrass theorem).

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E.g. $\{a+ib: a, b\in D\}$, where $D$ is any set, dense in $\mathbb{R}$. Some examples of $D$ are: $\mathbb{R}$ itself, $\mathbb{Q}+\sqrt{2}\mathbb{Q}$ or $\sqrt{3}\mathbb{Q}$.

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That is correct. But I think one needs some more interesting dense sets (if available) than these common sets. As this example: In $\mathbb{R}^+$, if $a,b>1$ are relatively prime integers,then $\left{\frac{a^n}{b^k}:n,k\in \mathbb{N} \right}$ is dense in $\mathbb{R}^+$. – Sara Oct 29 '14 at 09:26
A beautiful countable example dense in $\ \mathbb R^2\ $ was given by Wacław Siepiński. It is the graph of function
$$\ S:\ \mathbb Z\ +\ \sqrt 2\cdot\mathbb Z\ \ \rightarrow \ \ \mathbb Z+\sqrt 2\cdot\mathbb Z $$
where
$$ \forall_{a\ b\in\mathbb Z}\quad S(a+\sqrt 2\cdot b)\ :=\ b+\sqrt 2\cdot a $$
Wow!

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As Przemysław Scherwentk suggested earlier, all the sets defined as $\{a+ib \, \mid \, a,b\in D \} $, for some dense subset $D\subset\mathbb{R}$, are dense in $\mathbb C$.
If one is looking for some unsual example, even though the following is somewhat common, one can consider $D := (-\infty,0)\cup \mathcal A \cup (1,+\infty)$, where $\mathcal A$ is the complement of the Cantor set.