Find the complex numbers $z$ such that $w=\dfrac{2z-1}{2+iz}$ is real.
I have been trying to separate the imaginary part from the real one, so I can cancel the imaginary one. Trouble is that, by manipulating $w$, it just seems to get much worse.
Any path to follow?
HINT:
$$\frac{2z-1}{2+iz}=\frac{2z-1}{2+iz}\,\frac{2-i\bar z}{2- i\bar z}=\frac{4z-2-i2|z|^2-i\bar z}{4+|z|^2+4\text{Im}(z)}$$
Now, answer the question "what values of $z$ makes the numerator real?"
$$w=\frac{(2x-1)+iy}{(2-y)+ix}=\frac{[(2x-1)+iy][(2-y)-ix]}{(2-y)^2+x^2}$$
Multiply by the numerator to get
$$(2x-1)(2-y)-i(2x^2-x)-i(2y-y^2)-xy$$
The imaginary part is $x-2x^2+y^2-2y$. Setting it equal to zero, you can conplete the squares on $x$ and $y$ to maybe get a familiar geometric shape....
$$-2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right)+(y^2-2y+1)+\frac{1}{8}-1=0$$
$$(y-1)^2-2\left(x-\frac{1}{4}\right)^2=\frac{7}{8}$$
