If - z -4-2-2- then the maximum value of is A- -3-1B- -5-1C- -5-1D- -3-1

The correct option is B √(5)+1|z|= | z 4/2+4/z | ≤ | z 4/z |+ | 4/2 | = | z 4/z |+4/|z| |z|≤ 2 +4/|z| |z|^2 2|z|≤ 4 |z|^2 2 |z|+1 ≤ 5 ⇒ |z|≤√(5)+1 ...

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Byju's Answer

Standard XII

Mathematics

Inequality of 2 Complex Numbers

If | z -4/2|=...

Question

If $∣ ∣ z - \frac{4}{2} ∣ ∣ = 2,$ then the maximum value of is

\sqrt{3} + 1

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\sqrt{5} + 1

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\sqrt{5} - 1

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\sqrt{3} - 1

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Solution

The correct option is B $\sqrt{5} + 1$
$| z | = ∣ ∣ z - \frac{4}{2} + \frac{4}{z} ∣ ∣ \leq ∣ ∣ z - \frac{4}{z} ∣ ∣ + ∣ ∣ \frac{4}{2} ∣ ∣ = ∣ ∣ z - \frac{4}{z} ∣ ∣ + \frac{4}{| z |} | z | \leq 2 + \frac{4}{| z |} | z |^{2} - 2 | z | \leq 4 | z |^{2} - 2 | z | + 1 \leq 5 \Rightarrow | z | \leq \sqrt{5} + 1$

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The correct option is B √5+1|z|=∣∣z−42+4z∣∣≤∣∣z−4z∣∣+∣∣42∣∣=∣∣z−4z∣∣+4|z||z|≤2+4|z||z|2−2|z|≤4|z|2−2|z|+1≤5⇒|z|≤√5+1

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