Find the locus of $z$ such that $\arg \frac{z-z_1}{z-z_2} = \alpha$.
Use and draw $w = \frac{z-z_1}{z-z_2}$.
This exercise was discussed many times — 1, 2, 3, 4 — but I was unable to find answers to my problem with $0$ there.
I believe I understand where the arcs came from, here's my work:
If I understand correctly, for $\alpha = \pm\pi$, the locus would be the segment connecting $z_2$ and $z_1$, not including the points themselves.
I can not understand what is happening when $\alpha = 0$.
$\alpha = 0 = \arg \frac{z-z_1}{z-z_2} = \arg w \Longrightarrow \frac{z-z_1}{z-z_2} = k \in \mathbb{R}, \frac{z-z_1}{z-z_2} = k\frac{z-z_2}{z-z_2}.$
Solving this for $z$, $z = \frac{x_1-kx_2}{1-k} +i\frac{y_1-ky_2}{1-k}$. I am having trouble understanding the locus of this $z$. The textbook says it should be 'two segments with end points in $z_1$ and $z_2$, and one of this segments contains an infinitely distant point'. How to understand why is this answer right, and how to draw it? It seems the infinitely distant point matches $k=1$, but why should it lie in the 'direction' of the line passing through $z_1$ and $z_2$?
My class notes are messy. Why is $(0, 1)$ special on $w$ plane?
Thank you.
Best Answer
You know that (within a modulus of $2\pi$),
$\arg \left(w_1 \times w_2\right) ~=~ \arg(w_1) + \arg(w_2).$
Therefore, $\arg \left(\frac{w_1}{w_2}\right) ~=~ \arg(w_1) - \arg(w_2).$
In this situation, you have that
$$0 = \alpha ~=~ \arg \left(\frac{z - z_1}{z - z_2}\right) ~=~ \arg(z - z_1) - \arg(z - z_2).$$
Imagine the infinite line that passes through $z_1$ and $z_2$.
Note, that the $\arg$ function is not defined on the complex number $(0 + i[0]).$
Therefore, $z$ is not allowed to equal either $z_1$ or $z_2$.
There are 3 possibilities:
$\underline{\text{case 1} ~z ~\text{is not on this line}}$
Then,
$$\arg(z - z_1) \neq \arg(z - z_2).$$
Therefore, this possibility must be excluded from the locus of satisfying points.
$\underline{\text{case 2} ~z ~\text{is on this line}, ~\textbf{but between} ~z_1 ~\text{and} ~z_2}$
Then,
$$\arg(z - z_1) ~=~ \arg(z - z_2) ~\pm ~\pi.$$
Therefore, this possibility must also be excluded from the locus of satisfying points.
$\underline{\text{case 3} ~z ~\text{is on this line}, ~\textbf{but not between} ~z_1 ~\text{and} ~z_2}$
Then, regardless of whether the point $z$ is closer to $z_1$ or closer to $z_2$,
$$\arg(z - z_1) ~=~ \arg(z - z_2).$$
Therefore, this possibility represents the locus of all satisfying points.
Thus, the locus of all satisfying points, when $\alpha = 0,$ is all $z$ that are on the line formed by $z_1$ and $z_2$, but are not between $z_1$ and $z_2$.