Are there as many real numbers as there are imaginary numbers? On the one hand, I know that $\mathbb{R}$ and $\mathbb{I}=\{xi:x\in\mathbb{R}\setminus\{0\}\}$ are both uncountable sets,  so they have the same number of elements (i.e. the same cardinality) 
On the other hand, there's no bijection between $\Bbb{R}$ and $\Bbb{I}$: $0$ is not mapped to anything in $\Bbb{I}$, so, by definition, $\require{enclose} \enclose{horizontalstrike}{\mathbb{R}}$ and $\enclose{horizontalstrike}{\mathbb{I}}$ have different sizes (cardinalities) .
These two statements seem to contradict each other, so which one is correct?
Please excuse my ignorance and/or lack of correct terminology; I'm a rookie when it comes to set theory. 

Edit: the statements with strikeouts are erroneous and have later been shown to be nonsense, but I've included them for completeness.
 A: I can give you an explicit bijection from $\mathbb{R} \mapsto \mathbb{I^+}$.  Map $x \mapsto i e^x$.  
A: There is a bijection between $\Bbb R$ and $\Bbb C$. Therefore there is an injection from $\Bbb I$ into $\Bbb R$, and of course there is an injection from $\Bbb R$ into $\Bbb I$. By the Cantor-Bernstein theorem, there is a bijection between the two sets.
To see why there is a bijection between $\Bbb R$ and $\Bbb C$ it's very easy to note that there is a bijection between $\Bbb R$ and $\Bbb{N^N}$ (the set of infinite sequences of natural numbers), and then observe that: $$(\Bbb{N^N})^2\approx\Bbb{N^{2\times N}}\approx\Bbb{N^N}.$$
Therefore $\Bbb C$, which naturally has a bijection with $\Bbb R^2$, has the same cardinality as $\Bbb R$.

The question has been edited, and now it redefines $\Bbb I$ as the set $\{xi\mid x\in\Bbb R\setminus\{0\}\}$.
Here a bijection is easily definable. It is true that $x\mapsto xi$ is not a bijection since $0$ is causing us problems. There are two easy ways to solve this problem:


*

*$\Bbb I$ maps injectively into $\Bbb R$ by mapping $xi$ to $x$, obviously; and in the other direction $x\mapsto e^xi$ is an injection as well. Therefore $\Bbb R$ and $\Bbb I$ have the same cardinality. But we can do better, we can write down an explicit bijection.

*Note that only one element is causing us problems, so we just need to "shift" some elements around. For example:
$$x\mapsto\begin{cases} xi & x\notin\Bbb N\\(x+1)i & x\in\Bbb N\end{cases}$$ and in this context, $0\in\Bbb N$.
The key issue is that not every bijection needs to be "very simple" or even continuous. Or even definable by nice means.
