I have been looking at some complex number locci problems and I was wondering how would one intuitively think about them. The one I have in mind is what is the locus of $\arg{\frac{z-1}{z}} =\alpha$ given that $\alpha$ is such that $0<\alpha<\frac{\pi}{2}$. First, I thought of letting $z=a+bi$ and then see what the locus of $z$ is algebraically. I got that it is going to be a circle, however, when I checked the answer I was wrong as the answer was the circle above the real axis. Here is the image from the answers: 
My attempt at thinking about it visually
We have $\arg{\frac{z-1}{z}}=\arg{z-1}- \arg{z}= \alpha$. So we want to find the points on the argrand diagram s.t the angle between $z-1$ and $z$ is equal to $\alpha$. In trying to apply this diagrammatically, I ploted a general point $z$ and then $z-1$. I plotted $\alpha$ but from there I got stuck as how to proceed.
Could someone give me some insight into how to interpret the problem more visually and intuitively?
Thanks in advance.
Best Answer
As seen from the diagram, the inner angle $\alpha$ of the triangle is the difference between the outer angle $\arg(z-1)$ and the other inner angle $\arg(z)$, i.e.
$$\alpha = \arg(z-1) - \arg(z) = \arg\frac{z-1}z$$
For constant $\alpha$, the vertex of $\alpha$ follows the circumference of the red circle, but only above the $x$-axis. It can be shown similarly that the subtended angle below the $x$-axis is $\arg\frac{z}{z-1}\ne \alpha$, thus excluded.