<s>[INST] How to show that the number $w =\frac{ z}{1+z^2}$ is real only if $|z| = 1$ when $y\ne0$ 
If  $z = x + iy$  where $y \ne 0$,  show that the number $w = \frac{ z}{1+z^2}$    is real only if $|z| = 1 .$

Hi, so I've subbed $x+iy$ into the components into where $z$ is but where do I go from there?
Thanks.
 [/INST] $$\dfrac z{1+z^2}=\dfrac{x+iy}{1+x^2-y^2+i(2xy)}=\dfrac{(x+iy)(1+x^2-y^2-i2xy)}{(1+x^2-y^2)^2+(2xy)^2}$$
The imaginary part of $(x+iy)(1+x^2-y^2-i2xy)$
is  $$-2x^2y+y(1+x^2-y^2)=y(1-x^2-y^2)$$
</s>