Confusion about complex operations (stereographic projection)

Question

Hi I am reading the solution for the problem below and understand the general idea except the part where the write equates $X' to -X$. Why did the $|z|^{2}$ disappeared and why did $|\frac{1}{\bar{z}}|$ became $|z|^{2}$ ? What algebraic properties am I missing here?

I.3.2 If the point $P$ on the sphere corresponds to $z$ under the stereographic projection, show that the antipodal point $-P$ on the sphere corresponds to $-1 / \bar{z}$. Solution For $z=x+i y$ the corresponding point on the sphere under transformation given on p. 12 in CA is given by $(X, Y, Z)$ where $$ \begin{aligned} X &=\frac{2 x}{|z|^{2}+1}, \\ Y &=\frac{2 y}{|z|^{2}+1}, \\ Z &=\frac{|z|^{2}-1}{|z|^{2}+1} \end{aligned} $$ For $z \neq 0$, we have $$ -\frac{1}{\bar{z}}=-\frac{1}{x-i y}=-\frac{x+i y}{(x-i y)(x+i y)}=-\frac{x+i y}{x^{2}+y^{2}}=-\frac{x}{|z|^{2}}-i \frac{y}{|z|^{2}} $$ which corresponds to the point on the sphere given by $\left(X^{\prime}, Y^{\prime}, Z^{\prime}\right)$ where $$ \begin{aligned} X^{\prime} &=\frac{\frac{-2 x}{|z|^{2}}}{\left|\frac{1}{\bar{z}}\right|^{2}+1}=\frac{-2 x}{|z|^{2}+1}=-X, \\ Y^{\prime} &=\frac{\frac{-2 y}{|z|^{2}}}{\left|\frac{1}{\bar{z}}\right|^{2}+1}=\frac{-2 y}{|z|^{2}+1}=-Y, \\ Z^{\prime} &=\frac{\left|\frac{1}{\bar{z}}\right|^{2}-1}{\left|\frac{1}{\bar{z}}\right|^{2}+1}=-\frac{|z|^{2}-1}{|z|^{2}+1}=-Z \end{aligned} $$ Hence, the map $(X, Y, Z) \longmapsto(-X,-Y,-Z)$ on the sphere corresponds to $z \longmapsto-1 / \bar{z}$ of $\mathbb{C} .$

Answer 1

A complex rational expression (in elementary algebra, not to be confused with the "complex" from complex numbers) is a rational expression which contains rational expressions inside of it. Have you ever simplified complex rational expressions before? There are a couple ways to do it.

• Simplify the major numerator and denominator into a single fraction (which may require adding fractions) and then use $\frac{A/B}{C/D}=\frac{AD}{BC}$.

• Find the lowest common minor denominator and multiply the major numerator and denominators by it. In simplifying $X'$ for example, multiply the top and bottom of the fraction by $|z|^2$. The numerator becomes $-2x$ and the denominator becomes $1+|z|^2$.

Answer 2

Since $|z|=|\overline{z}|$, in the denominator of $X'$ we have $$ \bigg|\frac{1}{\overline{z}}\bigg|^2+1 = \frac{1}{|{\overline{z}}|^2}+1 = \frac{1}{|{z}|^2}+1 = \frac{|z|^2+1}{|z|^2}$$ which leads to $-X$.