How can it be true that $|z|^2 = 1$ for every complex number $z$?

Question

$$z=x+iy$$ $$\frac{z}{1+z}=\frac{z\bar{z}}{\bar{z}+z\bar{z}}=\frac{|z|^2}{|z|^2+\bar{z}}$$ $$z\bar{z}=|z|^2\implies\bar{z}=\frac{|z|^2}{z}$$ $$\Rightarrow \frac{|z|^2}{|z|^2+\bar{z}}=\frac{|z|^2}{|z|^2+\frac{|z|^2}{z}}=|z|^2\Bigg(\frac{1}{1+\frac{1}{z}}\Bigg)=|z|^2\Bigg(\frac{1}{\frac{z+1}{z}}\Bigg)=|z|^2\Bigg(\frac{z}{z+1}\Bigg)$$ $$\boxed{\therefore \Bigg(\frac{z}{1+z}\Bigg)=|z|^2\Bigg(\frac{z}{1+z}\Bigg)}$$

This implies that $|z|^2=1$. Everything checks out but the result seems strange, in the sense that is the following correct, then? $$|z|=1 ; \,x,y\in ℝ$$

Answer 1

This step is incorrect: $$\frac{|z|^2}{|z|^2+\frac{|z|^2}{z}}=|z|^2\Bigg(\frac{1}{1+\frac{1}{z}}\Bigg).$$ You have factored $|z|^2$ out of both the numerator and denominator, so the right side should be $$\frac{|z|^2}{|z|^2}\Bigg(\frac{1}{1+\frac{1}{z}}\Bigg)$$ instead.

(Also, there is an issue with division by $0$ if $z=0$ or $z=-1$, but this is minor.)

Answer 2

No, the second equality in the fourth line is not valid. If you remove $|z|^2$ from its right-hand side, it's okay.

Answer 3

your mistake lies on this line:

$\frac {|z|^2}{|z|^2 + \frac {|z|^2}{z}} = \frac {1}{1 + \frac {1}{z}}$

(at least if $z\ne 0$)