# Intuition behind simple complex number loci problems

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? 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.
• Also, I know this is a stupid question, but, why is $z-1$ intersecting at one as shouldn't the lines $z and$z-1$start from the origen and have their real coordinate being displaced by 1? – Maths Wizzard Apr 27 at 11:19 • @MathsWizzard - note that on a circle, any angle subtending a fixed chord, in this case from 0 to 1, is the same. So, given the constant$\alpha\$, the locus is along the red circle – Quanto Apr 27 at 14:49