Is there an image to the point at infinity through this map? I encountered a conformal mapping on the complex plane:$$z\rightarrow e^{i\pi z}$$
and I am not sure about where it does send the point at infinity. If I could say something along the lines: $$\text{Im}(\infty) = \infty$$
Then it would map it to the origin but there is still a voice in my head saying that this equality is non-sense. And from the usual definition of the imaginary part:$$\text{Im}(z)=\frac{z+\bar{z}}{2i}$$ it makes even less sense.
I'd appreciate some enlightenment.

You are seeking
$$
\lim_{z \to \infty} e^{i\pi z}.
$$
Note that if you consider $z = x + iy$, then by De Moivre's Theorem, we have
$$
e^{i\pi z} = e^{i\pi (x+iy)} = \frac{\cos(\pi x) + i \sin(\pi x)}{e^y}.
$$
Now everything depends on along which path $z \to \infty$: if $x$ is fixed and $y \to \infty$, the limit converges to $0$. Alternatively, if $y$ is fixed and $x \to \infty$, the limit does not exist.
Since the value of the limit is path-dependent, the limit does not exist.
