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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:

picture showing 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? messy class notes

Thank you.

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2 Answers 2

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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 $z_2$.

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  • $\begingroup$ 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. $\endgroup$ Commented Dec 30, 2020 at 6:09
  • $\begingroup$ @user2661923 In case two I think it should be $\arg(z - z_1) ~-~ \arg(z - z_2) ~\pm ~\pi.$ $\endgroup$
    – mathophile
    Commented Mar 19, 2023 at 4:10
  • $\begingroup$ @mathophile Your comment apparently has a typo, since your assertion contains no equal sign. If you intend the assertion that $$\arg(z - z_1) - \arg(z - z_2) ~\color{red}{=}~ \pm \pi,$$ then your assertion is equivalent to my statement that $$\arg(z - z_1) = \arg(z - z_2) \pm \pi.$$ If this is not what you are asserting, please re-state your assertion. $\endgroup$ Commented Mar 19, 2023 at 5:01
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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.

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  • $\begingroup$ I was going to answer the same $\endgroup$ Commented Nov 1, 2020 at 23:23
  • $\begingroup$ @raffaele But I think It won't be a complete circle but rather it will be a part of circle. Because argument will be zero only if number is positive real number $\endgroup$
    – mathophile
    Commented May 14 at 16:31

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