Q99 / 225 · JEE Main 2002 · Mathematics

If $|z-4| < |z-2|$, its solution is given by

  • A
    Re(z) > 0
  • B
    Re(z) < 0
  • Correct answer
    C
    Re (z) > 3
  • D
    Re(z) > 2

Solution

Given $|z-4| < |z-2|$ Let $z=x+$ iy $$ \begin{aligned} & \Rightarrow \mid(x-4)+i y)| < |(x-2)+i y \mid \Rightarrow(x-4)^2+y^2 < (x-2)^2+y^2 \\ & \Rightarrow x^2-8 x+16 < x^2-4 x+4 \Rightarrow 12 < 4 x \Rightarrow x>3 \Rightarrow \operatorname{Re}(z)>3 \end{aligned} $$

Hence, the correct answer is option: 'C'.

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