Find the complex numbers $z$ such that $w=\frac{2z-1}{2+iz}$ is real

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....

$$-2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right)+(y^2-2y+1)+\frac{1}{8}-1=0$$

$$(y-1)^2-2\left(x-\frac{1}{4}\right)^2=\frac{7}{8}$$