If -2z-1-z-2- and z1- z2- z3 are complex numbers such that -z1- - - -z2- - - then -z1-z2-

The correct option is B | 2z 1 |=| z 2 | or | 2z 1 |^2=| z 2 |^2 or (2z 1)(2z 1)=(z 2)(z 2) or 4zz 2z 2z+1=zz 2z 2z+4 or 3| z |^2=3 o ...

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Algebra of Complex Numbers

If |2z-1|=|...

Question

If $| 2 z - 1 | = | z - 2 |$ and $z_{1}, z_{2}, z_{3}$ are complex numbers such that $| z_{1} - α | < α, | z_{2} - β | < β$ , then $| \frac{z_{1} + z_{2}}{α + β} |$

< | z |

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> | z |

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> 2 | z |

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Solution

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Similar questions

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

z_{1}

z_{2}

z_{3}

| z_{1} - α | < α

| z_{2} - β | < β

∣ ∣ ∣ \frac{z_{1} + z_{2}}{α + β} ∣ ∣ ∣

{| z_{1} + z_{2} |}^{2} + {| z_{1} - z_{2} |}^{2} = {2 | z_{1} |}^{2} + 2 {| z_{2} |}^{2}

∣ ∣ ∣ α + \sqrt{α^{2} {- β}^{2}} ∣ ∣ ∣ + ∣ ∣ ∣ α - \sqrt{α^{2} - β^{2}} ∣ ∣ ∣ = | α + β | + | α - β |

z - 1, z_{2}

z_{3}

| z_{1} | = | z_{2} | = | z_{3} | = | (1 / z_{1}) + (1 / z_{2}) + (1 / z_{2}) | = 1

| z_{1} + z_{2} + z_{3} |

| z | = 2

\frac{z_{1} - z_{3}}{z_{2} - z_{3}} = \frac{z - 2}{z + 2}

z_{1}, z_{2}, z_{3}

| z | = 2

\frac{z_{1} - z_{3}}{z_{2} - z_{3}} = \frac{z - 2}{z + 2}

z_{1}, z_{2}, z_{3}

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