# 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.

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I am really confused by your question. What is the domain of your map? What does "$\Im (\infty) = \infty$" mean? What do you mean by "the point of infinity"? –  user79202 May 29 '13 at 14:32
This map has an essential singularity at $\infty$. Can you see that there are complex numbers with arbitrarily large magnitude whose imaginary parts are $0$, or which diverge to $-\infty$ (in the real number sense)? –  Erick Wong May 29 '13 at 14:37

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.

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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.

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The image of the map is not the unit circle, it's clearly the same as the image of $e^z$, which is $\mathbb C\setminus\{0\}$. –  Erick Wong May 29 '13 at 14:42
@ErickWong My bad. Got carried away with real arguments. Fixed. –  gt6989b May 29 '13 at 14:46