If -z1 - 1- - 1- -z2 - 2- - 2- -z3 - 3- - 3- then find the greatest value of -z1 - z2 - z3-

The correct option is D the greatest value is 12.Given, |z_1 + #160; #160;z_2 + z_3| = |(z_1 1) + (z_2 2) + (z_3 3) + 6| ≤|z_1 1| + |z_2| ...

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Definition of Relations

If |z1 - 1|...

Question

If $| z_{1} - 1 | \leq 1, | z_{2} - 2 | \leq 2, | z_{3} - 3 | \leq 3$ , then find the greatest value of $| z_{1} + z_{2} + z_{3} |$ .

the greatest value is 6.

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the greatest value is 7.

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the greatest value is 9.

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the greatest value is 12.

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Solution

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Similar questions

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

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

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

| 9 z_{1} z_{2} + 4 z_{1} z_{3} + z_{2} z_{3} | = 12,

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

If |z+7| ≤ 9, for z ∈ C, the greatest value of |Z+2| is __

| $Z_{1}$ + $Z_{2}$ | ≤ | $Z_{1}$ | + | $Z_{2}$ |

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

| 9 z_{1} z_{2} + 4 z_{1} z_{3} + z_{2} z_{3} | = 12

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

z_{1}

z_{2}

z_{3}

\frac{1}{∣ z_{1} - z_{2} ∣} = \frac{3}{∣ z_{2} - z_{3} ∣} = \frac{5}{∣ z_{1} - z_{3} ∣}

\frac{1}{z_{1} - z_{2}} + \frac{9}{z_{2} - z_{3}} + \frac{25}{z_{3} - z_{1}}

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