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 A $R e (z) > 0$
We know that, $| z - z_{1} | > | z - z_{2} |$ represents the region on right side of perpendicular bisector of $z^{1}$ and $z^{2}$ .
Thus, the given inequality $| z - 2 | > | z - 4 |$ represents the region on the right side of perpendicular bisector of $2$ and $4$ .
$\Rightarrow R e (z) > 3$ and $Im (z) \in R$

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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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