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Suppose $|a|<1$ and $r\in (0,1)$. Show that the set of complex number $z$ satisfying $\left|\dfrac{z-a}{1-\bar a z}\right|=r$ is a circle in complex plane. Find the centre and radius of this circle.

By putting $z=x+iy$ and $a=a_1+ia_2$ and then simplifying I got the result, but the process is too lengthy. I want a short and tricky way that the equation represents a circle.

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Starting from $$\left|\dfrac{z-a}{1-\bar a z}\right|=r$$ we can get $$\dfrac{\left|z-a\right|}{\left| z-1/{\bar a}\right|}=r\left|a\right|$$ which in language of geometry means 'point $z$ distances from two fixed points $a$ and $1/{\bar a}$ make a constant ratio (equal $r\left|a\right|$)'.

And that is a definition of a Circle of Apollonius (Wikipedia Circle, section Circle of Apollonius) — so you have a circle (provided the RHS is not equal $1$, in which case you get a line bisecting the segment with endpoints $a$ and $1/{\bar a}$).

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  • $\begingroup$ +1. As one acquires more mathematical tools, one should not forget about basic facts. $\endgroup$
    – mastrok
    Aug 24 '15 at 6:55
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$$\left|\dfrac{z-a}{1-\bar a z}\right|^2=r^2$$ $$(z-a)\overline{(z-a)}=r^2(z\bar{a}-1)\overline{(z\bar{a}-1)}$$ $$|z|^2-(z\bar a+\bar z a)+|a|^2=r^2(|z|^2|a|^2-(z\bar a+\bar z a)+1)$$ $$|z|^2(1-r^2|a|^2)-(z\bar a+\bar z a)(1-r^2)+(|a|^2-r^2)=0$$ Let $z=x+iy$ and $a=p+iq.$ Then this equation can be simplified to $$x^2+y^2-(px+qy)\left(\dfrac{1-r^2}{1-r^2(p^2+q^2)}\right)+\dfrac{p^2+q^2-r^2}{1-r^2(p^2+q^2)}=0.$$

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  • $\begingroup$ I did this calculation ...But from here how I can say that this represents a circle and how detect the centre and radius ? $\endgroup$
    – Empty
    Jul 16 '15 at 6:51
  • $\begingroup$ See my edit. That will helps you. $\endgroup$
    – Bumblebee
    Jul 16 '15 at 6:56
  • $\begingroup$ Oh!! Actually finally you put $z=x+iy$ and $a=p+iq$. I tried to find out the equation in the form $|z-a|=R$. $\endgroup$
    – Empty
    Jul 16 '15 at 7:00
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    $\begingroup$ Actually $$|z-a|=r\iff |z|^2-(z\bar a+\bar z a)+(|a|^2-r^2)=0.$$ Can you find $a$ and $r$ comparing above equations? I think it will not be difficult. $\endgroup$
    – Bumblebee
    Jul 16 '15 at 7:05

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