Discussion Forum

If the point $\left(\frac{k-1}{k},\frac{k-2}{k}\right)$ lies on the  locus of z satisfying  the inequality  $|\frac{z+3i}{3z+i}|$  <1, then the interval in which k lies is 

Answer:

Explanation:

Given ,

  $|\frac{z+3i}{3z+i}|$  <1

 $|\frac{x+(y+3)i}{3(x+iy)+i}|  <1$  [Let z=x+iy]

  | x+(y+3)i|  < | 3x+(3y+1)i|

 $\Rightarrow$   $x^{2} +(y+3)^{2}   <   9 x^{2}+(3y+1)^{2}$

$\Rightarrow$      $x^{2}+y^{2}+6y+9  < 9 x^{2}+9 y^{2}+6y+1$

$\Rightarrow$      $8x^{2}+8 y^{2}-8 >0$

 $\Rightarrow$       $x^{2}+y^{2}-1 >0$

 $\frac{k-1}{k}, \frac{k-2}{k}$ lie on locus of z

 $\therefore$       $\frac{(k-1)^{2}}{k^{2}}+\frac{(k-2)^{2}}{k^{2}}-1 >0$

 $\Rightarrow$    $k^{2}-2k+1+k^{2}-4k+4-k^{2} >0$

 $\Rightarrow$ $k^{2}-6k+5 >0 \Rightarrow  (k-5)(k-1) > 0$

 $\Rightarrow$    k ε  $(-\infty ,1) \cup (5, \infty)$