# A stereographic estimate relevant inquiry

This could be a very easy inquiry, yet I have not had the ability to up think of a remedy.

The photo 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 device circle eliminate the north post. $f$ includes a function $$g: \mathbb{C} \backslash \{i, -i \} \to \mathbb{C}^2.$$ Can anything be claimed concerning the photo of $g$?

0
2019-12-02 02:47:36
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Note that although $a$ is intricate, stands :

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

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

Thus is additionally an circle

EDIT

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

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2019-12-03 04:20:33
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It is the device circle. Replacing $\tan t$ for $a$ and also streamlining, you get

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

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2019-12-03 04:20:25
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