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: enter image description here

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.


enter image description here

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.

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  • $\begingroup$ Hi, thanks for the answer! Could you explain to me how one arrives that the locus will be a circle non-algebraically? $\endgroup$ – Maths Wizzard Apr 27 at 10:11
  • $\begingroup$ 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? $\endgroup$ – Maths Wizzard Apr 27 at 11:19
  • $\begingroup$ @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 $\endgroup$ – Quanto Apr 27 at 14:49
  • $\begingroup$ @MathsWizzard - regarding z-1, complex numbers can also be viewed as vectors in the Argend plane and z-1 is the difference between the vector z and 1, an arrow from 1 to z. $\endgroup$ – Quanto Apr 27 at 14:58
  • $\begingroup$ all clear now, thank you for taking the time to explain this to me. $\endgroup$ – Maths Wizzard Apr 27 at 16:13

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