= Solution
For the given <Möbius transformation>, the <fixed point> equation is $z(z+3)=3z+1$, hence
$$
\boxed{z=1,\qquad z=-1.}
$$
For the general <Möbius transformation>, its finite fixed points satisfy
$$
cz^2+(d-a)z-b=0.
$$
With $c\ne0$ and nonzero <discriminant>, the <quadratic formula> gives the two distinct <fixed points of a Möbius transformation>:
$$
\boxed{\alpha=\frac{a-d+m}{2c},\qquad\beta=\frac{a-d-m}{2c},\qquad m^2=(a-d)^2+4bc.}
$$
The fixed-point equation for $\alpha$ also gives $b-d\alpha=-\alpha(a-c\alpha)$. Therefore
$$
\omega-\alpha=\frac{(a-c\alpha)(z-\alpha)}{cz+d},
\qquad
\omega-\beta=\frac{(a-c\beta)(z-\beta)}{cz+d}.
$$
Dividing yields the <conjugation of a Möbius transformation with two distinct fixed points>:
$$
\boxed{\frac{\omega-\alpha}{\omega-\beta}
=k\frac{z-\alpha}{z-\beta},\qquad
k=\frac{a-c\alpha}{a-c\beta}=\frac{a+d-m}{a+d+m}.}
$$
Neither factor in this quotient vanishes, since $(a-c\alpha)(a-c\beta)=ad-bc\ne0$. The formula extends over the <Riemann sphere>, including the transformation's pole. For the special transformation and the ordering $\alpha=1,\beta=-1$, $k=1/2$.
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