# Fun with a sphere

1. Mar 13, 2006

### stunner5000pt

Show that the stereographic projections of the points z and $1/\overline{z}$ are reflections of each other in teh equatorial plane of the Reimann sphere

ok so let z = x + iy
then $$\frac{1}{\overline{z}} = \frac{x - iy}{x^2 + y^2}$$
so the magnitude of $$\frac{1}{\overline{z}}$$ is $$\frac{1}{x^2 + y^2}$$

the stereogrpahic projection of z is
$$x_{1} = \frac{2x}{x^2 + y^2 +1}$$
$$y_{1} = \frac{2y}{x^2 + y^2 +1}$$
$$z_{1} = \frac{x^2 + y^2 -1}{x^2 + y^2 +1}$$

for 1/ z bar is
$$x_{2} = \frac{2x(x^2 + y^2)^2}{1 + (x^2 + y^2)^2}$$
$$y_{2} = \frac{2y(x^2 + y^2)^2}{1 + (x^2 + y^2)^2}$$
$$z_{2} = \frac{1-(x^2 + y^2)^2}{1 + (x^2 + y^2)^2}$$
i fail to see how these are reflections of each other, then shouldnt the x1 = - x2?? and so on??

Last edited: Mar 13, 2006
2. Mar 13, 2006

### assyrian_77

3. Mar 13, 2006

### stunner5000pt

perhaps im missing something here??

4. Mar 13, 2006

### assyrian_77

Well, the magnitude of a complex number $z=x\pm iy$ is $|z|=\sqrt{x^2+y^2}$, but you probably know this already. You have the complex number

$$\frac{1}{\overline{z}}=\frac{x-iy}{x^2+y^2}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}$$

I get a different magnitude than yours.

5. Mar 13, 2006

### stunner5000pt

ys i got what u got ( i made a sign error)

so the magnitude would be...
$$\sqrt{\left(\frac{x}{x^2 + y^2}\right)^2 +\left(\frac{y}{x^2 + y^2}\right)^2} = \sqrt{\frac{1}{x^2 + y^2}}$$

the projection is then

$$x_{2} = \frac{2x(x^2 + y^2)}{1 + (x^2 + y^2)}$$

right?
i know sily math errors everywhere!!

6. Mar 14, 2006

### stunner5000pt

is what i did in the above post correct now??