Find all the complex numbers $z$ such that $w=\frac{2z-1}{z-2}$ has argument $\frac{\pi}2$

Question

I'm stuck here:

Find all the complex numbers $z$ such that $w=\frac{2z-1}{z-2}$ has a principal argument of $\frac{\pi}2$.

I don't see the way to solve it without having to expand this expression like this:

$$a+bi=\frac{2(x+yi)-1}{(x+yi)-2}$$

with the condition $a=0$. I think that this is not the best way to solve it, so I'd appreciate any help.

Thanks.

Answer 1

$$w=\frac{(2x-1+2yi)(x-2-yi)}{(x-2)^2+y^2},$$ which gives $$(2x-1)(x-2)+2y^2=0$$ and $$-(2x-1)y+2(x-2)y>0.$$ There are infinitely many these numbers.

Answer 2

You can solve it geometrically: consider the Argand-Cauchy plane. The condition is equivalent to $$\operatorname{Arg}\frac{z-\frac12}{z-2}=\frac\pi2,$$ and the set of images of $z$ which satisfy this relation is the circle with diameter the points of the real axis $\frac12$ and $2$, these points being removed.