<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] $$\dfrac Z{1+Z^2}=\dfrac{x+iy}{1+x^2-y^2+2xyi}$$
$$=\dfrac{(x+iy)(1+x^2-y^2-2xyi)}{(1+x^2-y^2)^2+(2xy)^2}$$
We need $y(1+x^2-y^2)-x(2xy)=0\implies y(1+x^2-y^2-2x^2)=0\iff x^2+y^2=1$
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