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I'm having trouble coming up with an easy-to-describe surjection $f : \mathbb{R} \to \mathbb{C}$. Here's what I came up with: (Edit: this doesn't work; see Qiaochu's comment)

Define $P(x \in \mathbb{R}) = \frac{1}{1-e^{-x}}$ and $P^{-1}(x \in (0,1)) = -\log \frac{1-x}{x}$.

Let $f(x) = P(\operatorname{even}(P^{-1}(x))) \cdot e^{2 \pi i \operatorname{odd}(P^{-1}(x))}$, where

$\operatorname{even}(0.b_0b_1b_2\ldots) = 0.b_0b_2b_4\ldots, \\\operatorname{odd}(0.b_0b_1b_2\ldots) = 0.b_1b_3b_5\ldots.$

However, this seems needlessly complicated. Is there a simpler way?

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    @Snowball That wouldn't add much to the existing answer and I did not describe what happens to negative numbers. It's not complete yet.2012-12-02

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Basically, what you need to do is find a way to produce from one real number a pair of real numbers (to get the imaginary and real parts of the image). A standard trick to obtain two real numbers from one is as follows. First fix some convention that chooses between two decimal expansions (or any other base (except base 1)) in case a real number has two decimal expansions. Now, given a real number $t\in [0,1)$ with its given decimal expansion $0.d_1d_2\cdots $ define $x=0.d_1d_3d_5\cdots$ and $y=0.d_2d_4d_6\cdots$. This gives a bijection $[0,1)\to [0,1)\times [0,1)$ which you can extend in many ways to get a surjection (even a bijection) $\mathbb R\to \mathbb C$.

Later addition: As noted in the comments, the construction is not quite a bijection and the interval [0,1) should be changed to [0,1]. This can be corrected by noting that when the construction is restricted to irrationals the image misses only countably many elements. These then can be paired up with the rationals since they are also countable.

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    It does not map to $[0, 1)^2$ and it is not a surjection to $[0,1]^2$ either since it misses $(1,1)$.2012-12-02
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Since $f:\ {\mathbb R}\to{\mathbb R}^2$ is only required to be surjective you can even construct a continuous $f$ doing the trick:

Take your favorite Peano curve in the unit square, i.e., a continuous surjective function $\phi:\ [0,1]\to[0,1]^2$. Put $\phi(0)=: a$, $\phi(1)=:b$.

Number the integer squares $[j,j+1]\times[k,k+1]$ with even numbers $2\ell$ going from $-\infty$ to $\infty$, such that $[0,1]^2$ becomes $Q_0$, and you have two symmetrically outward going "spirals" of sequentially adjacent squares $Q_2$, $Q_4$, $Q_6$, $\ldots$, resp., $Q_{-2}$, $Q_{-4}$, $\ldots$, covering the whole plane. Denote the lower left corner of $Q_{2\ell}$ by $z_{2\ell}$.

Then for each $\ell\in{\mathbb Z}$ use the $t$-interval $[2\ell, 2\ell+1]$ to cover $Q_{2\ell}$ with a Peano curve by putting $f(t):=z_{2\ell}+\phi(t-2\ell)\qquad(2\ell\leq t\leq 2\ell+1)\ .$ Finally for each $\ell\in{\mathbb Z}$ use the $t$-interval $[2\ell+1, 2\ell+2]$ to connect the endpoint of the Peano curve in $Q_{2\ell}$ with the initial point of the Peano curve in $Q_{2(\ell +1)}$: $f(t):=(2\ell+2-t)(z_{2\ell}+b)+(t-2\ell-1)(z_{2(\ell+1)}+a)\qquad(2\ell+1\leq t\leq 2\ell+2)\ .$