If |2z−1|=|z−2| and z1, z2, z3 are complex numbers such that z1−α<α,z2−β<α, then z1+z2α+β

|2z−1|=|z−2|or |2z−1|2=|z−2|2or (2z−1)(2z−1)=(z−2)(z−2)or 4zz−2z−2z+1=zz−2z−2z+4or 3|z|2=3or |z|=1Again,z1+z2=z1−α+z2−β+α+β≤z1−α+z2−β+|α+β| 0]∴ z1+z2α+β<2or z1+z2α+β<2|z|

If |2z−1|=|z−2| and z1, z2, z3 are complex numbers such that z1−α<α,z2−β<α, then z1+z2α+β

|2z−1|=|z−2|or |2z−1|2=|z−2|2or (2z−1)(2z−1)=(z−2)(z−2)or 4zz−2z−2z+1=zz−2z−2z+4or 3|z|2=3or |z|=1Again,z1+z2=z1−α+z2−β+α+β≤z1−α+z2−β+|α+β| 0]∴ z1+z2α+β<2or z1+z2α+β<2|z|

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If $| 2 z - 1 | = | z - 2 |$ and z₁, z₂, z₃ are complex numbers such that $|z_{1} - α| < α, |z_{2} - β| < α$ , then $|\frac{z_{1} + z_{2}}{α + β}|$

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< |z|

< 2|z|

> |z|

> 2|z|

answer is B.

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Detailed Solution

$| 2 z - 1 | = | z - 2 |$

$\begin{array}{l} or | 2 z - 1 |^{2} = | z - 2 |^{2} \\ or (2 z - 1) (2 z - 1) = (z - 2) (z - 2) \\ or 4 z z - 2 z - 2 z + 1 = z z - 2 z - 2 z + 4 \\ or 3 | z |^{2} = 3 \\ or | z | = 1 \end{array}$

Again,

$\begin{matrix} |z_{1} + z_{2}| = |z_{1} - α + z_{2} - β + α + β| \\ \leq |z_{1} - α| + |z_{2} - β| + | α + β | \\ < α + β + | α + β | \\ = 2 | α + β | [∵ α, β > 0] \end{matrix}$