3

Taking $z=x+iy$ and solving the equation algebraically shows that it is a circle but I cannot understand intuitively why this is so.

I would greatly appreciate a geometrical intuition behind this like there is with $|z-a|= r$ (postive numerical value), where $|z-a|$ represents the distance between $z$ and $a$, and $a$ being a fixed point, it represents a circle with radius $r$.

Also geometrical understanding of cases where $\arg\left(\frac{z-a}{z-b}\right)=\theta$ represent a pair of straight lines or other things. Thanks in advance!

Source: Problem-11 Tristan Needham Visual Complex Analysis page-46

tryst with freedom
  • 11,538
Roshan Surana
  • 31
  • Welcome to MSE. Please use MathJax. – José Carlos Santos Oct 02 '17 at 08:22
  • I think it represents a circle. The line joining $a,b$ can be treated as a chord of this circle. This equation necessarily means that you have to find a locus of points which subtend a constant angle on a line segment. Such a curve is circle. You have to verify whether its just an arc or a complete circle. – jonsno Oct 02 '17 at 08:23
  • https://math.stackexchange.com/a/2441754/108128 – Nosrati Oct 02 '17 at 10:11

2 Answers2

3

What is $\boldsymbol{\arg\left(\frac{z-a}{z-b}\right)}$?

$$ \begin{align} \arg\left(\frac{z-a}{z-b}\right) &=\arg\left(\frac{a-z}{b-z}\right)\\[3pt] &=\arg(a-z)-\arg(b-z) \end{align} $$ enter image description here

That is, $\arg\left(\frac{z-a}{z-b}\right)$ is the counterclockwise angle from $b$ to $a$ as viewed from $z$.

For a given $\boldsymbol{\theta}$, what is the locus of $\boldsymbol{\arg\left(\frac{z-a}{z-b}\right)=\theta}$ ?

The Inscribed Angle Theorem says that the locus of points at which the angle from $b$ to $a$ is $\theta$ is an arc of a circle containing $a$ and $b$ so that the angle from $b$ to $a$ at the center of the circle is $2\theta$:

enter image description here

robjohn
  • 345,667
  • Where did you find $$\begin{align} \arg\left(\frac{z-a}{z-b}\right) &=\arg\left(\frac{a-z}{b-z}\right)\[3pt] &=\arg(a-z)-\arg(b-z) \end{align}$$? It seems to be inconsistent with traditional identities for $\tan^{-1}x-\tan^{-1}y$. Just asking. – Cye Waldman Oct 03 '17 at 20:07
  • 1
    @CyeWaldman: $\arg(zw)=\arg(z)+\arg(w)\pmod{2\pi}$ – robjohn Oct 03 '17 at 21:56
  • 1
    Thanks. I see now how to derive that from $$\theta=\frac{1}{2i}\ln\left(\frac{z}{z^*}\right)$$. – Cye Waldman Oct 03 '17 at 22:36
0

$$\arg\frac{z-a}{z-b}=\theta$$ Recall that $\arg w=\arctan\frac{\Im w}{\Re w}$

If $a=p+qi;\;b=r+si$ then

$$\arg\frac{z-a}{z-b}=\arctan\frac{-p s+p y+q r-q x-r y+s x}{p r-p x+q s-q y-r x-s y+x^2+y^2}$$

Apply $\tan$ to both sides and get

$$\frac{-p s+p y+q r-q x-r y+s x}{p r-p x+q s-q y-r x-s y+x^2+y^2}=\tan\theta$$

Which can be rearranged and give the equation of a circle

$$x^2+y^2+hx+ky+j=0$$ where, called $\tan\theta=t$

$h= \left(-p+\frac{q}{t}-r-\frac{s}{t}\right)$

$k= \left(-\frac{p}{t}-q+\frac{r}{t}-s\right)$

$j=p r+\frac{p s}{t}-\frac{q r}{t}+q s$

Raffaele
  • 26,371