The inequality $| z - 4 | < | z - 2 |$ represents the following region

R e (z) > 0

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R e (z) < 0

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R e (z) > 2

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none of these

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Solution

The correct option is D none of these
$| z - 4 | < | z - 2 |$
Let $z = x + i y .$
Thus, we have | $x - 4 + i y | < | x - 2 + i y |$
$\Rightarrow \sqrt{x^{2} - 8 x + 16 + y^{2}} < \sqrt{x^{2} - 4 x + 4 + y^{2}}$
Squaring, we have $x^{2} - 8 x + 16 + y^{2} < x^{2} - 4 x + 4 + y^{2}$
$\Rightarrow 4 x > 12$ or $x > 3$ i.e. $R e (z) > 3$

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