Let $D=\{z\in\mathbb{C}|2<Re(z)<3\}$. Let $f(z)=\frac{z-3}{z-4}$. Find $f(D)$.
I thought of using the decomposition of Mobius transformations, i.e.
$$f_1(w)=w-4,\ f_2(w)=1/w,\ f_3(w)=1+w$$
Then we get
$$f(z)=\frac{z-3}{z-4}=\frac{z-4+1}{z-4}=1+\frac{1}{z-4}=f_3\Big(f_2\big(f_1(z)\big)\Big)$$
So $f(D)=f_3\Big(f_2\big(f_1(D)\big)\Big)$, and now we can deal with each transformation, i.e.
$$D_1=f_1(D)=\{z-4\in\mathbb{C}|2<Re(z)<3\}=\{z\in\mathbb{C}|-2<Re(z)<-1\}$$
And $f_3$ is similar to $f_1$. The problem is that I don't understand what the inversion $f_2$ will do on $D_1$.
Write $$w=\frac{z-3}{z-4}\implies z=\frac{4w-3}{w-1}$$
Now put $w=u+iv$ and simplify, so you get $$z=\frac{4u^2+4v^2-7u+3-iv}{(u-1)^2+v^2}$$
The region in question, $2<Re(z)<3$ becomes $$2<\frac{4u^2+4v^2-7u+3}{(u-1)^2+v^2}<3$$
The LHS simplifies to become $$(u-\frac 34)^2+v^2>\frac 74$$ The RHS simplifies to become $$(u-\frac 12)^2+v^2<\frac 14$$