I know how to find a transformation given three "regular" points, but I am getting confused as to how to handle this setup. I am given $T(1) = 4$, $T(0) = i$, and $T(\infty) = -1$.
I am using the cross ratio method, and obtain \begin{align*} \left( z, 1; 0, \infty \right) & = \left( w, 4; i, -1 \right) \\[10pt] \frac{z \left( 1 - \infty \right)}{z - \infty} & = \frac{5 \left( w - i \right)}{\left( w + 1 \right) \left( 4 - i \right)} \end{align*}
I am unsure on how to preceed past this point. How can I simplify and remove the infinity term to solve for $w$?
The formula $${(z_{1},z_{2};z_{3},z_{4})={\frac {(z_{3}-z_{1})(z_{4}-z_{2})}{(z_{3}-z_{2})(z_{4}-z_{1})}}}$$ holds only if all $z_k$ are finite. The case where one $z_k$ is equal to $\infty$ can be obtained by a limiting process, for example $${(z_{1},z_{2};z_{3},\infty)={\frac {z_{3}-z_{1}}{z_{3}-z_{2}}}} \, .$$
Or use that $$ (z_{1},z_{2};z_{3},z_{4}) = T(z_1) $$ where $T$ is the (unique) Möbius transformation with $T(z_2) = 1$,$T(z_3) = 0$, and $T(z_4) = \infty$.
Which means that the left-hand side of your equation is simply $$ ( z, 1; 0, \infty ) = z \, . $$