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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.

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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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