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

R e (z) \geq 0

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R e (z) < 0

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R e (z) > 0

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none of these

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Solution

The correct option is D none of these
One way of solving the above question is substituting z with $x + i y$

$∴ | x + i y - 4 | < | x + i y - 2 |$

Squaring the above modulus, we get

$x^{2} - 8 x + 16 + y^{2} < x^{2} - 4 x + 4 + y^{2}$

$\Rightarrow 8 x - 4 x > 16 - 4$

$\Rightarrow x > 3$ or $R e (z) > 3$

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Similar questions

| z - 4 | < | z - 2 |

∣ z - 4 ∣<∣ z - 2 ∣

z = x + i y \in C : | \begin{matrix} z \end{matrix} | - R e (z) \leq 1

z = \sqrt{3 + 4 i} + \sqrt{- 3 + 4 i}

\pm \frac{π}{4}, \pm \frac{3 π}{4}

\sqrt{- 1} = i

\sqrt{z} = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \sqrt{\frac{| z | + R e (z)}{2}} + i \sqrt{\frac{| z | - R e (z)}{2}} & , if B > 0 \sqrt{\frac{| z | + R e (z)}{2} - i \sqrt{\frac{| z | - R e (z)}{2}}} & , if B < 0 \end{matrix}

z = {(\frac{\sqrt{3}}{2} + \frac{i}{2})}^{5} + {(\frac{\sqrt{3}}{2} - \frac{i}{2})}^{5},

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