# Find the locus of $z$ such that $\arg \frac{z-z_1}{z-z_2} = \alpha$: confused when $\alpha = 0$.

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.

Let $$z=x+iy;\;z_1=x_1+iy_1;\;z_2=x_2+iy_2$$ $$\frac{z-z_1}{z-z_2}= \frac{x^2-x (x_1+x_2)+x_1 x_2+(y-y_1) (y-y_2)}{(x-x_2)^2+(y-y_2)^2}+i\frac{x (y_2-y_1)+x_1 (y-y_2)+x_2 (y_1-y)}{(x-x_2)^2+(y-y_2)^2}$$ $$\text{arg}\left(\frac{z-z_1}{z-z_2}\right)=\arctan\frac{{\frac{x (y_2-y_1)+x_1 (y-y_2)+x_2 (y_1-y)}{(x-x_2)^2+(y-y_2)^2}}}{{\frac{x^2-x (x_1+x_2)+x_1 x_2+(y-y_1) (y-y_2)}{(x-x_2)^2+(y-y_2)^2}}}=\\=\arctan\frac{x (y_2-y_1)+x_1 (y-y_2)+x_2 (y_1-y)}{x^2-x (x_1+x_2)+x_1 x_2+(y-y_1) (y-y_2)}$$

$$\text{arg}\left(\frac{z-z_1}{z-z_2}\right)=0\to x (y_2-y_1)+x_1 (y-y_2)+x_2 (y_1-y)=0$$ Rearrange $$x (y_2-y_1)+y (x_1-x_2)+x_2y_1-x_1y_2=0$$ which is the equation of a line.

In the general case, let $$a=\cot\alpha$$

$$x^2+y^2+x (-x_1-x_2+y_1 a -y_2 a )+y (-x_1 a +x_2 a -y_1-y_2)+x_1 x_2+x_1 y_2 a -x_2 y_1 a +y_1 y_2=0$$ $$x^2+y^2+px+qy+r=0$$ we get a circle.

Hope this can be useful.

• I was going to answer the same Nov 1, 2020 at 23:23

You know that (within a modulus of $$2\pi$$),
$$\arg \left(w_1 \times w_2\right) ~=~ \arg(w_1) + \arg(w_2).$$

Therefore, $$\arg \left(\frac{w_1}{w_2}\right) ~=~ \arg(w_1) - \arg(w_2).$$

I can not understand what is happening when $$\alpha = 0.$$

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 $$z_2$$.

• Thank you for your clear and perspicuous explanation. I wanted to apologize that it took so long to accept the answer -- real life happened, and I forgot. Sorry. Thank you. Dec 30, 2020 at 6:09