Is there an image to the point at infinity through this map?

Question

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.

Answer 1

If we allow $z$ to approach the point at infinity along the positive imaginary axis, we find that $e^{i\pi z}$ tends to $0$. Approaching along the negative imaginary axis, we find that $e^{i\pi z}$ gets big without bound. It turns out that we can make $e^{i\pi z}$ tend toward anything we like, just by allowing $z$ to approach the point at infinity along an appropriate path, and along some paths (such as the positive real axis) it won't tend toward anything at all. Thus, there's no nice way to extend the map to the point at infinity.

Answer 2

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.