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
$$\dfrac Z{1+Z^2}=\dfrac{x+iy}{1+x^2-y^2+2xyi}$$
$$=\dfrac{(x+iy)(1+x^2-y^2-2xyi)}{(1+x^2-y^2)^2+(2xy)^2}$$
We need $y(1+x^2-y^2)-x(2xy)=0\implies y(1+x^2-y^2-2x^2)=0\iff x^2+y^2=1$
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 $$ 1=z\bar{z} = |z|^2\implies |z| = 1 $$
If I started with your solution $$-2yx^2 +y + x^2 y - y^3 =0\implies y(x^2+y^2-1)=0$$
It's easier to work with $w=\dfrac{1+z^2}z=\dfrac1z+z$ (that doesn't change the claim).
Taking the imaginary part,
$$-\frac y{x^2+y^2}+y=y\left(1-\frac1{x^2+y^2}\right)=0,$$ i.e. $$x^2+y^2=1.$$
Or, in polar coordinates,
$$-\frac1r\sin(\phi)+r\sin(\phi)=0\implies r=1.$$
HINT:
Let $z=r(\cos A+i\sin A)$ where $r\ge0,A$ are real
As $y\ne0,\sin A\ne0$
$1+z^2=1+r^2\cos2A+i(r^2\sin2A)$
$\dfrac1{1+z^2}=\dfrac{1+r^2\cos2A-ir^2\sin2A}{(1+r^2\cos2A)^2+(r^2\sin2A)^2}$
$r(\cos A+i\sin A)\cdot(1+r^2\cos2A-ir^2\sin2A)$ $=\cdots+ir[\sin A(1+r^2\cos2A)-r^2\cos A\sin2A]=\cdots+ir[\sin A-r^2\sin(2A-A)]$
$=\cdots+ir\sin A(1-r^2)$
I tried to solve this with geometry and how the arithmetic operations on complex numbers can be interpreted geometrically.
That means $\varphi(z) = \varphi(1+z^2)$.
Here's a little sketch of both those complex numbers drawn as arrows in the complex plane:
You can see $z$, $z^2$, $1$ and $1+z^2$ in green. I also added the arguments of $z$ and $z^2$ in red, for no real reason (pun not intended) other than to remember that they have a constant ratio of $\varphi(z) = 2\varphi(z^2)$. So you can't do much about that, but $\varphi(1+z^2)$ is a whole 'nother story, because of the $+1$
The true goal is still to line up the green arrow (i.e. $1+z^2$) with $z$. In order to be lined up, one has to be derivable from the other by multiplying with a real number $a$. I guess this is called "being linear dependent", but am not sure. Anyway, the formula for that looks like that:
$$ \begin{align} 1+z^2 &= az\\ z^2 -az +1 &= 0\\ z_{1/2} &= \frac{a}{2} \pm \sqrt{\left(\frac{a}{2}\right) ^2-1}\\ \end{align} $$
This is kind of a good thing, because there's $\pm$, which means this is a two for one. And circles aren't functions because they need two lines. So maybe we are on to something here. Let's see if the above formula can be squeezed into being a circle.
You said $z= x+iy$ where $y \neq 0$ and that means there has to be an imaginary part, so whatever $a$ is (cannot be bothered to calculate it), the term under the square will be negative and by that forces $z_{1/2}$ to have an imaginary part.
$$ \begin{align} z_{1/2} &= \frac{a}{2} \pm \sqrt{(-1)\left(-\left((\frac{a}{2}\right) ^2+1\right)}\\ &= \frac{a}{2} \pm \sqrt{-1} \sqrt{-\left(\frac{a}{2}\right)^2+1}\\ &= \frac{a}{2} \pm i \sqrt{-\left(\frac{a}{2}\right)^2+1}\\ &= x+iy\\ \end{align} $$
Finally, what's $|z|$?
$$\begin{align} |z| = |z_{1/2}| &= \sqrt{x^2 + y^2}\\ &= \sqrt{\left(\frac{a}{2}\right)^2 + \left(\sqrt{-\left(\frac{a}{2}\right)^2+1}\right)^2}\\ &= \sqrt{\left(\frac{a}{2}\right)^2 -\left(\frac{a}{2}\right)^2+1}\\ &=1 \end{align}$$
I'm not sure if this is rigorous enough to be a proof, but you just asked to show it, so I hope this is ok and offers a bit of a different approach.
