Find the image under the stereographic projection ($x_1=\frac{z+\bar{z}}{1+|z|^2}, x_2=\frac{z-\bar{z}}{i(1+|z|^2)}, x_3=\frac{|z|^2-1}{1+|z|^2}$) of the following sets on the sphere:

(a) the lower hemisphere $x_3\le 0$

(b) the polar cap $\frac{3}{4}\le x_3\le1 $

(c) lines of latitude $x_1=\sqrt{1-x_3^2}\cos\theta$, $x_2=\sqrt{1-x_3^2}\sin\theta$, for $x_3$ fixed and $0\le\theta \le 2\pi$

(d) lines of longitude $x_1=\sqrt{1-x_3^2}\cos\theta$, $x_2=\sqrt{1-x_3^2}\sin\theta$, for $\theta$ fixed and $-1\le x_3 \le 1$

I got so far

(a) $x^2+y^2\le 1$ everything inside the unit circle in complex plane.

(b) $x^2+y^2\ge 7$ everything outside the circle in complex plane.

(c) stuck

(d) stuck

Any help would be greatly appreciated. Thanks in advance.

  • $\begingroup$ Please, recall precisely what definition you take of the stereographic projection (there are different conventions) $\endgroup$ – Jean Marie Sep 12 '16 at 6:17
  • $\begingroup$ @JeanMarie Thanks for that. $\endgroup$ – mint Sep 12 '16 at 6:23
  • $\begingroup$ Very little detail : $x_3$ instead of $x_1$ in the last place. $\endgroup$ – Jean Marie Sep 12 '16 at 6:48

For (c), the fact that $x_3$ is fixed tells you that $|z|$ is fixed. So you can solve for $|z| = r_0$ for some $r_0$ in terms of your fixed $x_3$ and that will be your solution. Note that $x_1$ and $x_2$ can still take on the full range of values created by varying $\theta$, because you can vary the values of

$$y = (z - \overline{z})/2i \quad \text{and} \quad x = (z + \overline{z})/2$$

in accordance with the $\sin(\theta)$ and $\cos(\theta)$ dependence in $x_1$ and $x_2$, while still keeping $x^2 + y^2 = r_0^2$ constant.

For (d), note that $\tan(\theta) = x_2 / x_1 = y/x$ where $z = x + iy$, so fixing $\theta$ as defined in your equation also fixes $\theta$ in the complex plane. This means that your projection will be a line, but the fact that

$$ -1 \leq x_3 \leq 1$$

means that the value of $|z|$ is restricted. So you will get a radial segment in the plane as your projection

An easy way to get intuition for this is to note that those formulas for the stereographic projection give equations for the point on the unit sphere (which you've labeled as $(x_1, x_2, x_3)$) if you draw a line through the north pole of the sphere (i.e. $(x_1, x_2, x_3) = (0, 0, 1)$) and the point in the plane

$$z = x + iy \mapsto (x_1, x_2, x_3) = (x, y, 0) $$

see here

  • $\begingroup$ is it right what I did in (a) and (b)? $\endgroup$ – mint Sep 12 '16 at 8:36
  • $\begingroup$ Yes it looks so. For (a) and (b), you only restrict $x_3$, so you get the equations $|z| \leq 1$ for (a) and $|z|^2 \geq 7$ for (b), which is the same as what you wrote. $\endgroup$ – J. Marx-Kuo Sep 12 '16 at 17:05

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