Locus of circle using complex numbers

Question

I was going through Locus of circle using complex numbers and got stuck on one particular manipulation. We know that $|z-z_1|=r$ represents a circle whose centre is $z_1$ and radius r. Now in the book it goes further as follows: from $|z-z_1|=r$

$|z-z_1|^2=r^2$

$(z-z_1)(\bar z-\bar{z_1})=r^2$

$z\bar z -z\bar{z_1}-\bar zz_1+ z_1\bar{z_1}-r^2=0$ Let $\;-a=z_1 $ and $\;z_1\bar{z_1}-r^2=b\;$ where $ b\;\epsilon\; R $ .

Therefore equation of circle is $\;z\bar z +a\bar z + \bar az +b=0 $

Centre of circle is '-a' and radius is $\sqrt{a\bar a -b }.$

I cannot understand why the author use such a manipulation. I mean the previous form is more understandable. Why did he convert the locus into this form.??

Answer 1

The author did so for the following reasons You see in coordinate geometry eqn of a circle is $$x²+y²+2gx+2fy+c=0$$ The centre is $$(-g,-f)$$ And the radius is $$\sqrt{g²+f²-c}$$ So to create a resemblance he set $$z1=-a=g+if$$ And radius will be $$\sqrt{|a|²-c}$$

Answer 2

There is another reason connected to the important (non-linear) transformation called inversion, which, in its simplest form, is

$$z \mapsto Z=\frac{1}{\overline{z}} \iff z=\frac{1}{\overline{Z}}\tag{1}$$

(please note the involutivity of this transformation)

There is an important theorem saying that, by this transformation, a circle not passing through the origin is transformed into another circle not passing through the origin. How can you prove this result ? Easily, by using the given form :

$$z\bar z +a\bar z + \bar az +b=0\tag{2}$$

where be is a non-zero real number (indeed, if $b=0$, the circle passes through the origin).

Using (1) in (2), we get :

$$\frac{1}{\overline{Z}}\frac{1}{Z}+a\frac{1}{Z}+\bar a \frac{1}{\overline{Z}}+b=0$$

Using a common denominator $\frac{1}{Z\overline{Z}}$, we get :

$$1+a \bar Z+\bar a Z + b Z \bar Z=0$$

Otherwise said :

$$Z \bar Z+(a/b)\overline{Z}+\overline{(a/b)}Z+(1/b)=0$$

(because $b$ is a non-zero real number) which is indeed, again, a circle equation with :

$$\text{center} -a/b \ \ \text{and radius} \ \ \sqrt{|a/b|^2-(1/b)} $$

Remark : as a reminder of the fact that inversion is nonlinear : unlike for example homothety, the center of the image circle is not in general the image of the center which is $-1/\bar a$.