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How do we show that $|\frac{z-2}{z-3}|=2$ represents a circle, where $z\in\mathbb{C}$.

$$ \frac{(x-2)^2+y^2}{(x-3)^2+y^2}=4\implies 3x^2+3y^2-20x+32=0\implies(x-10/3)^2+y^2=(2/3)^2 $$ the substitution $z=x+iy$, shows that it represent a circle with center $(10/3,0)$ and radius $2/3$.

Is there another way to solve it without making the substitution ?

or is there any way to get an intuition in to the solution from a geometric consideration ?

For example:

If it was $|z-3|=2$,

we know that $|z-3|$ is the distance between $z$ and $-3+0i$, thus we can visualize it to be a circle with center $(-3,0)$ and of radius $2$.

Sooraj S
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2 Answers2

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Apollonius Circle (the first type) - Wolfram/Wikipedia

$(2, 0)$ and $(-3, 0)$ are fixed points in the Complex plane. The ratio of their distances from any solution, $z\in\Bbb C$, is $2$ (ie. constant) iff $\frac{|z-2|}{|z-3|}=\left|\frac{z-2}{z-3}\right|=2$. Hence the solution set is a circle.

Generalising, we can conclude equations of the form

$$\left|\frac{z-a}{z-b}\right|=r,\quad z\in\Bbb C$$

with constants $a,b\in\Bbb C, a\neq b$ and $r\in\Bbb R^+, r\neq 1$ have solution sets which are circles.

Shuri2060
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  • thnx for the hint Apollonius Circle, which I haven't heard before. I think it makes sense now, though i'm still trying to read more about it. – Sooraj S Aug 14 '17 at 16:17
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Yes, remember that $|z|^2 = z\cdot z'$

Write $|z-2 |^2=4|z-3|^2$.

Then we have $(z-2)(z'-2) =4(z-3)(z'-3)$.

After some calculation we get:

$$z\cdot z'-10/3 z -10/3 z' +32/3 =0$$

So $(z-10/3)(z'-10/3) -100/9 =-96/9$

Finnaly we have $|z-10/3|=2/3$

nonuser
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