Cardinality, surjective, injective function of complex variable. Let $A = \{ z\in \mathbb{C} : \operatorname{mod}(z)  \ge 1 \}$ and $B= \{ z\in \mathbb{C} : z \ne 0\} $. 
Does there exists an injective function from $A$ to $B$ or $B$ to $A$? 
Does there exist a surjective function from $A$ to $B$ or $B$ to $A$? 
I think it depends on cardinality of $A$ and $B$, but how to determine I am not sure. Does inversion function $f(z)=\dfrac1z$ may be useful in someway?
Thank you.
 A: You are right, this problem is related to cardinality: Two sets A and B are said to be of same cardinality if and only if a bijection between them exists.
For your problem, all four desired functions exist. To show that, there are at least three different approaches:


*

*You can show that an injection $f_1:A\to B$ and an injection $g_1:B\to A$ exist. From $f_1$ you can conclude $|A|\leq|B|$ and from $g_1$ you can conclude $|B|\leq|A|$. So $|A|=|B|$ and a bijection $h:A\to B$ exists. $h^{-1}:B\to A$ then is a bijection, too.

*You can show that a surjection $f_2:A\to B$ and a surjection $g_2:B\to A$ exist. From $f_2$ you can conclude $|B|\leq|A|$ and from $g_2$ you can conclude $|A|\leq|B|$. So $|A|=|B|$ and a bijection $h:A\to B$ exists. $h^{-1}:B\to A$ then is a bijection, too.

*You can show that a bijection $f:A\to B$ exists. $f^{-1}:B\to A$ then is a bijection, too.


We proceed by approach 1:


*

*Example for an injective function from $A$ to $B$:
$$f_{1}:=\begin{cases}
A & \to B\\
z & \mapsto z
\end{cases}$$

*Example for an injective function from $B$ to $A$:
$$g_{1}:=\begin{cases}
B & \to A\\
r\cdot e^{i\varphi} & \mapsto(r+1)\cdot e^{i\varphi}
\end{cases}$$
You can directly give a bijective function (approach 3), too:
$$f:=\begin{cases}
B & \to A\\
z & \mapsto\frac{\left|z\right|+1}{\left|z\right|}z\quad\text{(if }z\neq1\text{)}\\
z & \mapsto z^{2}\quad\text{(if }z=1,\arg(z)\in\left[0,\pi\right)\text{)}\\
z & \mapsto2z^{2}\quad\text{(if }z=1,\arg(z)\in\left[\pi,2\pi\right)\text{)}
\end{cases}$$
(EDIT 1: Corrected the answer.
EDIT 2: Added approach 3.)
A: Look at the sets geometrically:


*

*set $A$ is a unit circle plus its exterior,

*set $B$ is the whole complex plane wihtout the zero point,


As $A\subset B$, the identity function injects $A$ into $B$.
On the other hand you can inject $B$ into $A$ just by shifting each point away from the $0$: 
$$z\mapsto z\cdot\frac{|z|+1}{|z|}$$
Note, however, this function is not surjective: its image is an exterior of the unit circle, without the circle itself.
By the Schröder–Bernstein theorem, when there exist two injections $A\to B$ and $B\to A$, then there exists a bijection between the two sets, hence both sets are equipollent (and obviously there exists a surjection: the bijection is a surjection).
