Discussion Forum

A complex number z is said to be unimodular, if |z| =1, suppose  z₁ and z₂ are complex numbers such that $\frac{z_{1}-2z_{2}}{z-z_{1}z_{2}}$  is unimodular and z₂ is not unimodular.   Then , the point z₁ lies on a

Answer:

Explanation:

Central Idea:

if z is unimodular, then |z|=1, Also , use property of modulus i.e,   $z\overline{z}=|z|^{2}$

Given,   $z_{2}$  is not unimodular i.e, |$z_{2}$|≠ 1

 and   $\frac{z_{1}-2z_{2}}{2-z_{1}\overline{z_{2}}}$ is unimodular.

$\Rightarrow$    $\mid\frac{z_{1}-2z_{2}}{2-z_{1}\overline{z_{2}}}\mid$  =1

$\Rightarrow$      $\mid z_{1}-2z_{2}\mid ^{2}=\mid2-z_{1}\overline{z_{2}}\mid ^{2}$

$\Rightarrow$           $(z_{1}-2z_{2})(\overline{z_{1}}-2\overline{z_{2}})=(2-z_{1}\overline{z_{2}})(2-\overline{z_{1}}z_{2})$                                             $(\because  z\overline{z}= |z|^{2})$

$\Rightarrow$     $|z_{1}|^{2}+4|z_{2}|^{2}-2\overline{z_{1}}z_{2}-2z_{1}\overline{z_{2}}$

$\Rightarrow$     $(|z_{2}|^{2}-1)(|z_{2}|^{2}-4)=0$

  $\because$      $|z_{2}|\neq 1$

   $\because$   $|z_{1}|= 2$

 Let   $z_{1}=x+iy\Rightarrow x^{2}+y^{2}=(2)^{2}$

 $\therefore$    point z₁ lies on a circle of radius  2.