I had this simple statement:
$$\arg\left(\frac{z-1}{z}\right) = \frac{\pi}{2}$$ I was interested in the locus of $z$.
What my understanding is angle between the line segment joining $\overline{OA}$ and $\overline {OB}$ is $\frac{\pi}{2}$ where $O$ is origin, $A$ is $(x-1, y)$, and $B$ is $(x,y)$. Using "$\text{product of slopes} = -1$", I get the locus of $z$ as $x^2+y^2-x=0$, which is a circle centered at $(\frac{1}{2},0)$ and passing through the origin. But intuitively speaking, only $(\frac{1}{2}, -\frac{1}{2})$ should be a solution, and no other coordinate should satisfy.
What am I not considering here?
Let's first consider solving it directly. Then discuss some point you miss.
Since $\arg\left(\dfrac{z-1}{z}\right)=\dfrac{\pi}{2}$, this means $\dfrac{z-1}{z}\in\mathbb{R}^+i$, this means there is $k\in\mathbb{R}^+$ such that $$\dfrac{z-1}{z}=ki\iff z=\dfrac{1}{1-ki}=\dfrac{1+ki}{1+k^2}$$ So what is the locus of this set? You can see $(x,y)=\left(\dfrac{1}{1+k^2},\dfrac{k}{1+k^2}\right),$ we have $k=\dfrac{y}{x}$, so $$x=\dfrac{1}{1+\frac{y^2}{x^2}}\implies x^2+y^2=x;\quad x,y>0$$ i.e. the upper semicircle excluding the origin.
The point you are missing is that $z$ is lying on that semicircle, but $z-1$ is not. Possibly you can guess that $z-1$ is lying on the upper semicircle $x^2+y^2=-x$, and they have the same $y$-coordinate. So all points on the upper semicircle is possible (excluding the $x$-axis).