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.
 A: $$\dfrac z{1+z^2}=\dfrac{x+iy}{1+x^2-y^2+i(2xy)}=\dfrac{(x+iy)(1+x^2-y^2-i2xy)}{(1+x^2-y^2)^2+(2xy)^2}$$
The imaginary part of $(x+iy)(1+x^2-y^2-i2xy)$
is  $$-2x^2y+y(1+x^2-y^2)=y(1-x^2-y^2)$$
A: $$\frac {z}{z^2+1}\in \Bbb R $$
$$\iff \frac {z^2+1}{z}\in \Bbb R $$
$$\iff z+ \frac {1}{z}\in \Bbb R $$
$$\iff x+iy+\frac {x-iy}{x^2+y^2}\in \Bbb R $$
$$\iff y(x^2+y^2-1)=0$$
$$\iff x^2+y^2=1$$
Done!
A: HINT:
Note that we can write
$$\begin{align}
1/w&=z+1/z\\\\
&=x+iy +\frac{x-iy}{x^2+y^2}\\\\
&=\left(x+\frac{x}{x^2+y^2}\right)+i\left(y-\frac{y}{x^2+y^2}\right)
\end{align}$$
Now set the imaginary part to zero.
A: A complex number is real iff it is its own conjugate.  Applying that to $z/(1+z^2)$ quickly yields to 
$$(z-\bar z)(z\bar z-1)=0.$$
Since $z$ is not a real number, $z\bar z$, the squared length of $z$, must be one.
A: If $y=0$ then $z$ is real and therefore $w$ is real even if $|z|\neq 1$
if $y\not = 0$
$$w=\frac{z}{1+z^2}$$
Divide and multiply by $1+\overline z ^2$ you have
$$w=\frac{z(1+\overline z ^2)}{(1+\overline z ^2 +z ^2 +|z|^2)}$$
now $\overline z ^2 = \overline {z^2}$ and $z^2 + \overline {z}^2$ is therefore real.  Thus $w$ is real if and only if $z(1+\overline z ^2)$ 
$$z(1+\overline z^2) = z+|z|\overline z$$
Now write $z=(x+iy)$ then $z+|z|\overline z = x+|z|x+i(y-|z|y)$ this is real if and only if $y-|z|y=0$ since $y\neq 0$ it follows that $|z|=1$
