If |z3+1z3|≤2, then |z+1z| cannot exceed

Let |z+1z|=aNow, the identity,z3+1z3=(z+1z)3−3(z+1z) gives us |z+1z|3≤ |z3+1z3|+3|z+1z| ⇒a3≤2+3a⇒3a⇒a3−3a−2≤0 ⇒(a−2)(a+1)2≤0⇒a−2≤0 ⇒a≤2

If |z3+1z3|≤2, then |z+1z| cannot exceed

Let |z+1z|=aNow, the identity,z3+1z3=(z+1z)3−3(z+1z) gives us |z+1z|3≤ |z3+1z3|+3|z+1z| ⇒a3≤2+3a⇒3a⇒a3−3a−2≤0 ⇒(a−2)(a+1)2≤0⇒a−2≤0 ⇒a≤2

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If $| z^{3} + \frac{1}{z^{3}} | \leq 2, t h e n | z + \frac{1}{z} |$ cannot exceed

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$\sqrt{2} - 1$

$\sqrt{2}$

answer is A.

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Detailed Solution

$L e t | z + \frac{1}{z} | = a$

Now, the identity,

$z^{3} + \frac{1}{z^{3}} = {(z + \frac{1}{z})}^{3} - 3 (z + \frac{1}{z}) g i v e s u s {| z + \frac{1}{z} |}^{3} \leq | z^{3} + \frac{1}{z^{3}} | + 3 | z + \frac{1}{z} | \Rightarrow a^{3} \leq 2 + 3 a \Rightarrow 3 a \Rightarrow a^{3} - 3 a - 2 \leq 0 \Rightarrow (a - 2) {(a + 1)}^{2} \leq 0 \Rightarrow a - 2 \leq 0 \Rightarrow a \leq 2$