So I have these four transformations given by: $$z \mapsto z - \frac i 2$$
$$z \mapsto -z $$
$$z \mapsto \frac{z-i}{z+i}$$
$$z \mapsto z+1 $$
And I need to combine them into a single transformation. But every way I tried seems to get a different answer to the solution given.
The solution given was
$$z \mapsto \frac{4z-2i}{2z-3i}$$
On
$$f(z) = \begin{pmatrix}a &b \\ c & d\end{pmatrix} = \frac{az + b}{cz + d},~ z\rightarrow -z \\ \frac{a(-z) + b}{c(-z) + d} = \frac{-az + b}{-cz + d} = \begin{pmatrix} -a & b \\ -c & d\end{pmatrix}$$ so our transformation can be found by linear algebra to be $$\begin{pmatrix}a & b \\ c & d\end{pmatrix}\times \begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}-a & b\\ -c & d\end{pmatrix}$$ So the $z\rightarrow -z$ transform is given by the matrix $\begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix}$.
Since matrix multiplication represents such a transform, and matrix multiplication is associative, if we know all our transforms, then we can simply find the total transform by the resulting matrix product. Note that since in general matrix multiplication is not commutative, the order in which we apply our transforms is important.
$$z \mapsto z - \frac i 2 = \frac{2z - i}{2} = \frac{2z + (-i)}{0z + 2} \implies \begin{pmatrix}2 &-i \\ 0 & 2\end{pmatrix}$$
$$z \mapsto -z = \frac{-1z + 0}{0z + 1} \implies \begin{pmatrix}-1 &0 \\ 0 & 1\end{pmatrix}$$
$$z \mapsto \frac{z-i}{z+i} = \frac{1z + (-i)}{1z + i} \implies \begin{pmatrix}1 &-i \\ 1 & i\end{pmatrix}$$
$$z \mapsto z+1 = \frac{1z + 1}{0z + 1} \implies \begin{pmatrix}1 &1 \\ 0 & 1\end{pmatrix}$$
Multiply the matrices together (in the reverse order as stated):
$$\begin{pmatrix}1 &1 \\ 0 & 1\end{pmatrix}\begin{pmatrix}1 &-i \\ 1 & i\end{pmatrix}\begin{pmatrix}-1 &0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}2 &-i \\ 0 & 2\end{pmatrix} = \begin{pmatrix}-4 &2i \\ -2 & 3i\end{pmatrix}$$
Which corresponds to a translation of $\frac{-4z + 2i}{-2z + 3i} = \frac{4z - 2i}{2z - 3i}$, which matches your result.