Question
If $$\left| {z - 4} \right| < \left| {z - 2} \right|,$$ its solution is given by
A.
Re$$(z)$$ > 0
B.
Re$$(z)$$ < 0
C.
Re$$(z)$$ > 3
D.
Re$$(z)$$ > 2
Answer :
Re$$(z)$$ > 3
Solution :
$$\eqalign{
& {\text{Given}}\,\left| {z - 4} \right| < \left| {z - 2} \right|{\text{Let}}\,z = x + iy \cr
& \Rightarrow \,\,\left| {\left( {x - 4} \right) + iy)} \right| < \left| {\left( {x - 2} \right) + iy} \right| \cr
& \Rightarrow \,\,{\left( {x - 4} \right)^2} + {y^2} < {\left( {x - 2} \right)^2} + {y^2} \cr
& \Rightarrow \,\,{x^2} - 8x + 16 < {x^2} - 4x + 4 \cr
& \Rightarrow \,\,12 < 4x \cr
& \Rightarrow \,\,x > 3 \cr} $$
$$ \Rightarrow \,\,\operatorname{Re} \left( z \right) > 3$$