A Möbius map is a rational function of the form $$z\mapsto \frac{az+b}{cz+d} \,,$$ where $ad-bc\ne0$ and $a,b,c,d \in \mathbb{C}$.
Show that Möbius functions can be uniquely decomposed as $f=R\circ L \circ T$, where $R$ corresponds to a rigid rotation of the sphere, $L(z)=\alpha z$ for some $\alpha>0$, and $T(z)=z+\beta$, where $\beta\in \mathbb{C}$. (Hint: first consider the case $f(\infty)=\infty$; the general case can be reduced to this case by postcomposing $f$ with a rigid rotation.)
The question didn’t really tell me what a rigid rotation of sphere is, by googling I found a rigid rotation is $f(z)=\gamma z$ where $|\gamma|=1$.
Starting with hint, assume $f(\infty)=\infty$, then $c=0$, so what I’ve been doing is, set $f(z)=(az+b)/d$, and then try to solve the question using simple algebra. For example, I was thinking of picking $\alpha = 1/d$ and $\beta=b$. Then rotate $z/d \mapsto az/d$. But then $\alpha>0$ means it's a real number where $1/d$ is not necessarily a real number.
Let $f$ be an arbitrary Möbius transformation, and let $f(\infty) = P$ in the Riemann sphere. If $P=\infty$, then set the rotation $R$ equal to the identity.
If not, first note that if we have two distinct points $P$ and $Q$ on the unit sphere, there is a circle's worth of rotations of the sphere sending $P$ to $Q$. (You can see this either geometrically or with basic linear algebra. The rotations of the sphere must map oriented orthonormal basis to oriented orthonormal basis. Once we assign the value on the first basis vector, the value on the second basis vector is free to be in any circle orthogonal to that value. Assigning this determines the rotation uniquely.) In particular, since $Q=\infty$ is fixed by multiplication by $e^{i\theta}$, if $\rho$ is a rotation of the sphere carrying $P=f(\infty)$ to $\infty$, then all such possible rotations are given by $e^{i\theta}\rho$ for some $\theta$.
Consider, then, the Möbius transformation $\rho^{-1}\circ f$. Since it sends $\infty$ to $\infty$, it must be of the form $z\mapsto az+b$ for some $a,b\in\Bbb C$, $a\ne 0$. (As you observed, we must have $c=0$, and you can absorb the nonzero constant $d$ into $a$ and $b$ by dividing.) There is a unique $\theta$ so that $a=\alpha e^{i\theta}$ for some positive real number $\alpha$. Then $az+b = e^{i\theta}\alpha z + b = e^{i\theta}\alpha(z+\beta)$ for $\beta = b/a$.
Finally, set $R=e^{i\theta}\rho$, $L(z)=\alpha z$, and $T(z) = z+\beta$, with $\rho$, $\alpha$, and $\beta$ as already prescribed. Then $\rho^{-1}\circ f = e^{i\theta}L\circ T$, and so $f = R\circ L\circ T$, and all three functions are uniquely determined.