Using complex conjugates, one can show in a painfully long way that
$$|4z - 7| = 1 \iff \Re \left( \frac 1 {2-z} \right) = 2$$
Is there some way to show that this holds without using complex conjugates?
I have just tried something different. We want $$i\lambda = \frac {2z-3} {z-2}$$ to be pure imaginary. Then, up to a multiplicative constant $\mu$, we have $$ \newcommand \mat [1] {\begin{bmatrix} #1 \end{bmatrix}} \mat {i\lambda \\ 1} = \mu \mat {2 & -3 \\ 1 & -2} \mat {z \\ 1} $$ Inverting this system, we have $$ -\mu \mat {z \\ 1} = -\mat {2 & -3 \\ 1 & -2}^{-1} \mat {i\lambda \\ 1} = \mat {-2 & 3 \\ -1 & 2} \mat {i\lambda \\ 1} = \mat {3 - 2i\lambda \\ 2 - i\lambda} $$ Hence $\lambda$ is the parameter of the curve $$z = \frac {3 - 2i\lambda} {2 - i\lambda}$$ And I think I should be able to take it off from here. Wish me luck!
Here's how I did it. I'm not sure if it's the same way you have in mind or not. For simplicity, write $w=2-z$ so we want to show that $|1-4w|=1$ is the same as $\Re\left(\frac1w\right)=2$. The curve $|1-4w|=1$ is the circle of radius $\frac14$ about $w=\frac14$. Three points on the curve are $0,\ \frac12,$ and $\frac{1+i}4$. The transformation $w\mapsto\frac1w$ carries these points to $\infty,$ $2,$ and $2-2i$, and we are done.