# A stereographic projection related question

This might be an easy question, but I haven't been able to up come up with a solution.

The image of the map $$f : \mathbb{R} \to \mathbb{R}^2, a \mapsto (\frac{2a}{a^2+1}, \frac{a^2-1}{a^2+1})$$

is the unit circle take away the north pole. $f$ extends to a function $$g: \mathbb{C} \backslash \{i, -i \} \to \mathbb{C}^2.$$ Can anything be said about the image of $g$?

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The same as in the real case: it's the set of points satisfying x^2 + y^2 = 1, minus the point (0, 1). – Qiaochu Yuan Dec 7 '10 at 3:13
A difference is that the functions $a\mapsto\frac{2a}{a^2+1}$ and $a\mapsto\frac{a^2-1}{a^2+1}$ are no longer bounded. The first has image $\mathbb{C}$, while the second has image $\mathbb{C}\setminus\{1\}$. As Qiaochu says, the image of your map is still the set of pairs $(w,z)\in\mathbb{C}^2\setminus{(0,1)}$ such that $w^2+z^2=1$, and you can still represent such pairs using (complex) cosine and sine. – Jonas Meyer Dec 7 '10 at 3:39

Note that although $a$ is complex, is valid :

$$\left(\frac{2a}{a^2+1}\right)^2+\left(\frac{a^2-1}{a^2+1}\right)^2= \frac{4a^2}{(a^2+1)^2}+\frac{(a^2-1)^2}{(a^2+1)}=$$

$$\frac{4a^2+a^4-2a^2+1}{(a^2+1)^2}=\frac{(a^4+2a^2+1)}{(a^2+1)^2}=\frac{(a^2+1)^2}{(a^2+1)^2}=1$$

Thus is also an circle

EDIT

Is say, the points of the set $\{g(a)\in \mathbb{C}^2 :a\in \mathbb{C}/ \{\imath,-\imath\}\}$ meet the above.

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Yes, it is true that the sum of the squares is still 1, and Qiaochu commented that conversely each such pair of complex numbers other than $(0,1)$ is represented this way, but is there a good reason to call the resulting set a "circle"? – Jonas Meyer Dec 7 '10 at 3:29

It is the unit circle. Substituting $\tan t$ for $a$ and simplifying, you get

$$x=\sin 2t, y=-\cos 2t$$ which is the unit circle.

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I don't see a compelling reason to call this set the unit circle when you're using complex coordinates instead of real ones. – Qiaochu Yuan Dec 7 '10 at 3:45
You are right. What is the name of that set then? – TCL Dec 7 '10 at 3:52
Maybe the complex circle. As a Riemann surface, it's the Riemann sphere minus three points. (Of course, so is the LHS.) – Qiaochu Yuan Dec 7 '10 at 3:59
Perhaps a good way to view it is as a (co)tangent bundle of the unit circle (minus the point (0,1)). In fancy language they are Liouville isomorphic, and in a sense passing to cotangent bundle is the "right" complexification. – Max Dec 7 '10 at 15:33