Prove complex numbers $a$ and $b$ are antipodal under stereographic projection $\iff a \overline{b} = -1$ I'm trying to prove the following statement:

Given $a, b \in \mathbb{C}$, prove that $a$ and $b$ correspond to antipodal points on the Riemann sphere under stereographic projection if and only if $a \overline{b} = -1$


My attempt
I wanted to make a proof where all my implications were reversible to avoid making a proof of each implication separately. As previous knowledge, I know that if a have a point $a \in \mathbb{C}$, then the stereographic projection $f: \mathbb{C} \to S^2$ is given by
$$
f(a) = \left(\frac{a + \overline{a}}{1 + |a|^2},\frac{a - \overline{a}}{i\left(1 + |a|^2\right)},\frac{|a|^2-1}{|a|^2+1}\right)
$$
Now, given that $P,Q\in S^2$ are antipodal if and only if $P =-Q$, I get the following:
\begin{align}
f(a) = -f(b) &\iff
\begin{cases}
\frac{a + \overline{a}}{1 + |a|^2} = \frac{-b - \overline{b}}{1 + |b|^2} \\
\frac{a - \overline{a}}{i\left(1 + |a|^2\right)} = \frac{\overline{b}-b}{i\left(1 + |b|^2\right)} \\
\frac{|a|^2-1}{|a|^2+1} = \frac{1-|b|^2}{|b|^2+1} \\
\end{cases}\\
&\iff\begin{cases}
a + \overline{a}+a|b|^2 +\overline{a}|b|^2 = -b - \overline{b}-b|a|^2 -\overline{b}|a|^2 \\
a - \overline{a}+a|b|^2 -\overline{a}|b|^2 = -b + \overline{b}-b|a|^2 +\overline{b}|a|^2 \\
|ab|^2+|a|^2-|b|^2-1 =-|ab|^2+|a|^2-|b|^2+1 \\
\end{cases}\\
&\iff\begin{cases}
a +a|b|^2 = -b -b|a|^2  \\
\overline{a} +\overline{a}|b|^2 = -\overline{b} -\overline{b}|a|^2  \\
|ab|^2=1 \\
\end{cases}\\
&\iff\begin{cases}
a +b +a|b|^2+b|a|^2 =0 \\
|a||b|=1 \\
\end{cases}\\
\end{align}
Where here I use brackets to indicate that all those equations are true simultaneously. On this last step is where I ran into trouble because I couldn't find a way to show that both conditions in the last step are equivalent to $b =- \frac{1}{\overline{a}}$.
Is my attempt correct (up to what I have already written)? And if so, does somebody know how I could conclude the proof of equivalence? Any help would be greatly appreciated. Thank you!

For the direct implication, one could also use the inverse function of $f$, $\phi$:
$$ \phi (x,y,u) = \frac{x+iy}{1-u}$$
for $(x,y,u)\not= (0,0,1)$, $x^2+y^2+u^2=1.$
If $ P = (x,y,u)$ and $Q=(-x,-y,-u)$, then
$$ \phi(P)\overline{\phi(Q)} =  \frac{x+iy}{1-u} \cdot \frac{-x+iy}{1+u}  = -\frac{x^2+y^2}{1-u^2} = -1$$
The indirect implication is straightforward. For example:
$$\frac{a + \bar{a}}{1+|a|^2} =   \frac{-\bar{b}^{-1} -b^{-1}}{1+|b|^{-2}} = -\frac{b + \bar{b}}{1+|b|^2}.$$
Edit: Note that
$$ a + b + a|b|^2  + b|a|^2 = 0$$
is equivalent to
$$ a(1+|b|^2) = - b(1+|a|^2) $$
Multiplying by $\bar{b}$, we get:
$$ a\bar{b}(1+|b|^2) = - |b|^2(1+|a|^2) $$
which implies that $ a\bar{b}$ is real negative.
