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Determinant

If z1 = 5 +...

Question

If $z_{1} = 5 + 12 i$ and $| z_{2} | = 4$ , then

A

maximum

(| z_{1} + i z_{2} |) =

17

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B

maximum

(| z_{1} + (1 + i) z_{2}) = 13 + 4 \sqrt{2}

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C

minimum

∣ ∣ ∣ ∣ \frac{z_{1}}{z_{2} + \frac{4}{z_{2}}} ∣ ∣ ∣ ∣ = \frac{13}{4}

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D

minimum

∣ ∣ ∣ ∣ \frac{z_{1}}{z_{2} + \frac{4}{z_{2}}} ∣ ∣ ∣ ∣ = \frac{13}{5}

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Solution

The correct options are
A maximum $(| z_{1} + i z_{2} |) =$ 17
B maximum $(| z_{1} + (1 + i) z_{2}) = 13 + 4 \sqrt{2}$
D minimum $∣ ∣ ∣ ∣ \frac{z_{1}}{z_{2} + \frac{4}{z_{2}}} ∣ ∣ ∣ ∣ = \frac{13}{5}$
$z_{1} = 5 + 12 i & | z_{2} | = 4$
$| z_{1} + i z_{2} | \leq | z_{1} | + | i z_{2} | \Rightarrow | z_{1} + i z_{2} | \leq | 5 + 12 i | + | i | | z_{2} | \Rightarrow | z_{1} + i z_{2} | \leq 13 + 4 ∴ m a x i m u m (| z_{1} + i z_{2} |) = 17$
$| z_{1} + (1 + i) z_{2} | \leq | z_{1} | + | (1 + i) z_{2} | \Rightarrow | z_{1} + (1 + i) z_{2} | \leq | 5 + 12 i | + | i + 1 | | z_{2} | \Rightarrow | z_{1} + i z_{2} | \leq 13 + 4 \sqrt{2} ∴ m a x i m u m (| z_{1} + (1 + i) z_{2} |) = 13 + 4 \sqrt{2}$
$∣ ∣ ∣ \frac{z_{2}}{z_{1}} + \frac{4}{z_{1} z_{2}} ∣ ∣ ∣ \leq ∣ ∣ ∣ \frac{z_{2}}{z_{1}} ∣ ∣ ∣ + ∣ ∣ ∣ \frac{4}{z_{1} z_{2}} ∣ ∣ ∣ \Rightarrow ∣ ∣ ∣ \frac{z_{2}}{z_{1}} + \frac{4}{z_{1} z_{2}} ∣ ∣ ∣ \leq \frac{4}{| 12 + 5 i |} + \frac{4}{4 | 12 + 5 i |} = \frac{4}{13} + \frac{1}{13} \Rightarrow ∣ ∣ ∣ \frac{z_{2}}{z_{1}} + \frac{4}{z_{1} z_{2}} ∣ ∣ ∣ \leq \frac{5}{13} \Rightarrow ∣ ∣ ∣ ∣ ∣ ∣ \frac{z_{1}}{z_{2} + \frac{4}{z_{2}}} ∣ ∣ ∣ ∣ ∣ ∣ \geq \frac{13}{5} ∴ m i n i m u m ⎛ ⎜ ⎜ ⎜ ⎝ ∣ ∣ ∣ ∣ ∣ ∣ \frac{z_{1}}{z_{2} + \frac{4}{z_{2}}} ∣ ∣ ∣ ∣ ∣ ∣ ⎞ ⎟ ⎟ ⎟ ⎠ = \frac{13}{5}$
Hence, options 'A', 'B' and 'D' are correct.

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Similar questions

z_{1} = 5 + 12 i

| z_{2} | = 4

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| z_{1} | = 20

| z_{2} - 5 - 12 i | = 4

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