I am trying to solve a problem which states:
Let $\phi: \bar{\mathbb C} \to S^2$ be the inverse function of the stereographic projection Calculate $\phi(Re(z))=0$ and $\phi(Im(z))=0$.
I can guess that $\phi(Re(z))=0$ is a circle in the $yz$ plane and the other one a circle in the $xz$ plane. I've tried to formally prove that the respective images are those but I couldn't.
I know that if $z=a+ib$, then $\phi(z)=(\dfrac{2Re(z)}{1+|z|^2},\dfrac{2Im(z)}{1+|z|^2},\dfrac{|z|^2-1}{1+|z|^2})$.
Take the first case, $\phi(Re(z))=0$ is equal to the points of the form $(0,\dfrac{2b^2}{1+b^4},\dfrac{b^4-1}{1+b^4})$. I also know that these points must lie in $S^2$, so $||(0,\dfrac{2b^2}{1+b^4},\dfrac{b^4-1}{1+b^4})||^2=1$
From these equation I got $b \in \{-1,0,1\}$, that doesn't make any sense.
What am I doing wrong and what could I do to properly describe the respective images?
You performed the squaring earlier than it should have been used:
We have $Im(z)=b$ and $|z|^2=b^2$, and you took their squares.
With these, we get $\left(0,\displaystyle\frac{2b}{1+b^2},\frac{b^2-1}{1+b^2}\right)$, and we get it for any $b\in\Bbb R$.
(Check that its norm will be always $1$).