If $| z_{1} - 1 | < 1, | z_{2} - 2 | < 2, | z_{3} - 3 | < 3,$ then $| z_{1} + z_{2} + z_{3} |$

is less than 6

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is more than 3

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is less than 12

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lies between 6 and 12

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Solution

The correct option is C is less than 12
$| z_{1} - 1 | < 1$
$\Rightarrow - 1 < z_{1} - 1 < 1$
$\Rightarrow - 1 + 1 < z_{1} < 1 + 1$
$\Rightarrow 0 < z_{1} < 2$

$| z_{2} - 2 | < 2$
$\Rightarrow - 2 < z_{2} - 2 < 2$
$\Rightarrow - 2 + 2 < z_{2} < 2 + 2$
$\Rightarrow 0 < z_{2} < 4$

$| z_{3} - 3 | < 3$
$\Rightarrow - 3 < z_{3} - 3 < 3$
$\Rightarrow - 3 + 3 < z_{3} < 3 + 3$
$\Rightarrow 0 < z_{3} < 6$

$z_{1} + z_{2} + z_{3} < 2 + 4 + 6$
$\Rightarrow z_{1} + z_{2} + z_{3} < 12$
$∴ | z_{1} + z_{2} + z_{3} | < 12$

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Similar questions

if $z_{1}, z_{2} a n d z_{3}$ are complex numbers such that

| $z_{1}$ |=| $z_{2}$ |=| $z_{3}$ |= $∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} ∣ ∣ = 1$ ,

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

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

| z_{1} | = | z_{2} | = | z_{3} | = ∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} ∣ ∣ = 1

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

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

| z_{1} | = | z_{2} | = | z_{3} | = ∣ ∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} ∣ ∣ ∣ = 1

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

If $z_{1}, z_{2} a n d z_{3}$ are complex numbers such that

| $z_{1}$ |=| $z_{2}$ |=| $z_{3}$ |= $∣ ∣ \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} ∣ ∣ = 1$ ,

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

z_{1} = 1 + 2 i, z_{2} = 2 + 3 i, z_{3} = 3 + 4 i

z_{1}, z_{2}

z_{3}

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