What does $e^{i \pi \mathbf x} \in \mathbb C^2$ mean?

Question

I saw this in an book (coding theory) without explanation and I need help figuring out the meaning.

Consider $\mathbf x\in \{0,1\}^2$ and $$ \mathbf s = e^{i \pi \mathbf x} \in \mathbb C^2 \tag 1 $$

I know $\{0,1\}^2$ and $\mathbb C^2$ are the cartesian products, i.e. \begin{align} \{0,1\}^2&=\{0,1\}\times \{0,1\}\\ &=\{(0,0),(0,1),(1,0),(1,1)\} \tag 2 \\ \mathbb C^2 &=\mathbb C \times \mathbb C = \{(z,y): z,y\in \mathbb C\} \tag 3 \end{align} So $\mathbf x= (x_1,x_2)$ is a vector which take a value in the set $\{0,1\}^2$, for instance, $\mathbf x = (1,1)$.

So far so good, but what is $e^{i \pi \mathbf x} \in \mathbb C^2$? Any examples?

Say $\mathbf x = (1,1)$, what does $e^{i \pi (1,1)} \in \mathbb C^2$ mean?

Any clarification would be appreciated.

Answer 1

$e^{0} = 1$ and $e^{i\pi} = -1$ implying for example $e^{i \pi (1,1)} = e^{ (i \pi,i \pi)} = (-1,-1)$

so if $\mathbf x \in \{(0,0),(0,1),(1,0),(1,1)\}=\{0,1\}^2$

then $e^{i \pi \mathbf x} \in \{(1,1),(1,-1),(-1,1),(-1,-1)\} = \{1,-1\}^2 \subset \mathbb C^2$ and $e^{i \pi \mathbf x} \in \mathbb C^2$

Answer 2

In general you can without comment extend any function $f: A \to B$ to a function of the same name $f: A^n \to B^n$ by defining $f(a_1, a_2, ..., a_n) = (f(a_1), f(a_2), ... f(a_n)).$

In your particular case, $A = \left\{0, 1\right\}$, $B = \mathbb{C}$, $f(a) = e^{i \pi a}$, and $n = 2$.

So $e^{i \pi (x_1, x_2)} = (e^{i \pi x_1}, e^{i \pi x_2})$.

A simpler way to write exactly the same transformation is $f(x) = (-1)^x$. Basically the idea is to send all the $0$ bits to $1$ and send all the $1$ bits to $-1$.