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Complex Numbers

z 1 z 2 ∈ C, ...

Question

$z_{1} z_{2} \in C, {z_{1}}^{2} + {z_{2}}^{2} \in R, z_{1} ({z_{1}}^{2} - 3 {z_{2}}^{2}) = 2$ and $z_{2} (3 {z_{1}}^{2} - {z_{2}}^{2}) = 11$ , then the value of ${z_{1}}^{2} + {z_{2}}^{2}$ is

A

2

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B

3

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C

4

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D

5

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Solution

The correct option is D $5$
We have, $z_{1} ({z_{1}}^{2} - 3 {z_{2}}^{2}) = 2$ ...(1)
and, $z_{2} (3 {z_{1}}^{2} - {z_{2}}^{2}) = 11$ ...(2)
Multiplying (2) by $i$ and subtracting it from (1),w e get
${z_{1}}^{3} - 3 {z_{2}}^{2} z_{1} + i (3 {z_{1}}^{2} z_{2} - {z_{2}}^{3}) = 2 + 11 i$
$\Rightarrow {(z_{1} + i z_{2})}^{3} = 2 + 11 i$ ...(3)
Again, multiplying (2) by $i$ and subtracting it from (1), we get
${z_{1}}^{3} - 3 {z_{2}}^{2} z_{1} + i (3 {z_{1}}^{2} z_{2} - {z_{2}}^{3}) = 2 - 11 i$
$\Rightarrow {(z_{1} - i z_{2})}^{3} = 2 - 11 i$ ...(4)
Multiplying (3) and (4), we get
${({z_{1}}^{2} + {z_{2}}^{2})}^{3} = 4 + 121 \Rightarrow {z_{1}}^{2} + {z_{2}}^{2} = 5$

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Similar questions

z_{1} z_{2} \in C, z_{1}^{2} + z_{2}^{2} ϵ R, z_{1} (z_{1}^{2} - 3 z_{2}^{2}) = 2

z_{2} (3 z_{1}^{2} - z_{2}^{2}) = 11

, then the value of

z_{1}^{2} + z_{2}^{2}

\frac{a}{| z_{1} - z_{2} |} = \frac{b}{| z_{2} - z_{3} |} = \frac{c}{| z_{3} - z_{1} |} where (a, b, c \in R),

\frac{a^{2}}{z_{1} - z_{2}} + \frac{b^{2}}{z_{2} - z_{3}} + \frac{c^{2}}{z_{3} - z_{1}}

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

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

then the value of

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

If | $z_{1} | = | z_{2} | = | z_{3} | = | z_{4} | = 1$ and $z_{1} + z_{2} + z_{3} + z_{4} = 0$
then least value of the expression
$E = | z_{1} - z_{2} |^{2} + | z_{2} - z_{3} |^{2} + | z_{3} - z_{4} |^{2} + | z_{4} - z_{1} |^{2}$ is

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

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

then the value of

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

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