The inequality |z- 4| < |z- 2| represents region given by

Re(z) > 0

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Re(z) < 3

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Re(z) > 3

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Re(z) < 0

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Solution

The correct option is C Re(z) > 3
$| z - 4 | < | z - 2 |$
Substituting $z = x + i y$ , we have $| x - 4 + i y | < | x - 2 + i y |$
$\Rightarrow \sqrt{x^{2} - 8 x + 16 + y^{2}} < \sqrt{x^{2} - 4 x + 4 + y^{2}}$
$\Rightarrow x^{2} - 8 x + 16 + y^{2} < x^{2} - 4 x + 4 + y^{2}$
$\Rightarrow x > 3$
So, the region is given by $R e (x) > 3.$

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