JEE MAIN - Mathematics (2002 - No. 37)
If $$\left| {z - 4} \right| < \left| {z - 2} \right|$$, its solution is given by :
$${\mathop{\rm Re}\nolimits} (z) > 0$$
$${\mathop{\rm Re}\nolimits} (z) < 0$$
$${\mathop{\rm Re}\nolimits} (z) > 3$$
$${\mathop{\rm Re}\nolimits} (z) > 2$$
Explanation
Given $$\left| {z - 4} \right| < \left| {z - 2} \right|$$
Let $$\,\,\,z = x + iy$$
$$ \Rightarrow \left| {\left. {\left( {x - 4} \right) + iy} \right)} \right| < \left| {\left( {x - 2} \right) + iy} \right|$$
$$ \Rightarrow {\left( {x - 4} \right)^2} + {y^2} < {\left( {x - 2} \right)^2} + {y^2}$$
$$ \Rightarrow {x^2} - 8x + 16 < {x^2} - 4x + 4$$
$$ \Rightarrow 12 < 4x$$
$$ \Rightarrow x > 3$$
$$ \Rightarrow {\mathop{\rm Re}\nolimits} \left( z \right) > 3$$
Let $$\,\,\,z = x + iy$$
$$ \Rightarrow \left| {\left. {\left( {x - 4} \right) + iy} \right)} \right| < \left| {\left( {x - 2} \right) + iy} \right|$$
$$ \Rightarrow {\left( {x - 4} \right)^2} + {y^2} < {\left( {x - 2} \right)^2} + {y^2}$$
$$ \Rightarrow {x^2} - 8x + 16 < {x^2} - 4x + 4$$
$$ \Rightarrow 12 < 4x$$
$$ \Rightarrow x > 3$$
$$ \Rightarrow {\mathop{\rm Re}\nolimits} \left( z \right) > 3$$
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