<s>[INST] solving difficult complex number proving if $z= x+iy$ where $y \neq 0$ and $1+z^2 \neq 0$, show that the number $w= z/(1+z^2)$ is real only if $|z|=1$
solution :
$$1+z^2 = 1+ x^2 - y^2 +2xyi$$ 
$$(1+ x^2 - y^2 +2xyi)(1+ x^2 - y^2 -2xyi)=(1+ x^2 - y^2)^2 - (2xyi)^2$$
real component
$$(1+ x^2 - y^2)x - yi(2xyi) = x + x^3 + xy^2$$
imaginary component 
$$-2yx^2 i +yi + x^2 yi - y^3 i=0i$$
$$-2yx^2 +y + x^2 y - y^3 =0$$
... 
can't solve this question 
 [/INST] A another approach is to utilise (as @mann and myself concluded) the fact that 
$$
w = \bar{w}
$$ 
where bar represents the complex conjugate.
$$
\frac{z}{1+z^2} = \frac{\bar{z}}{1+\bar{z}^2}
$$
re-arrange to yeild
$$
z + z\bar{z}^2 = \bar{z} + \bar{z}z^2 \implies z-\bar{z} = \bar{z}z^2-z\bar{z}^2=z\bar{z}\left(z-\bar{z}\right)
$$
thus we get
$$
1=z\bar{z} = |z|^2\implies |z| = 1
$$
</s>