lf $| z_{1} | = 2, | z_{2} | = 3$ , then $| z_{1} + z_{2} + 5 + 12 i |$ is less than or equal to

8

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18

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10

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5

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Solution

The correct option is B $18$
We know that $| z_{1} + z_{2} + z_{3} | \leq | z_{1} | + | z_{2} | + | z_{3} |$
$| z_{1} | + | z_{2} | = 5$
$z_{3} = 5 + 12 i$
$| z_{3} | = 13$
$∴ 18 \geq | z_{1} + z_{2} + 5 + 12 i |$
Hence, option B is correct

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Similar questions

| z_{1} | = 2, | z_{2} | = 3,

| z_{1} + z_{2} + 5 + 12 i |

z_{1} = 5 + 12 i

| z_{2} | = 4

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

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