If |z|

1.	If \|z\|
Answer» Correct Answer - Option 1 : less than 2 Concept: Triangle inequality: \| z1 + z₂ \| \(\leq\)\| z₁\| + \| Z₂\| \| cos α \| \(\leq\) 1 Calculations: Given, \| z2 + 2z cos α \| \(\leq\)\| z2 \| + \| 2z cos α \| We know that, \| cos α \| \(\leq\) 1 ⇒ \| z2 + 2z cos α \| \(\leq\)\| z2 \| + \| 2z \| Put the value of \|z\| as √3 - 1 ⇒ \| z2 + 2z cos α \| < (√3 - 1)²+ 2 (√3 - 1) ⇒ \| z2 + 2z cos α \| < 2 Hence, If \|z\| < √3 - 1, then \|z2 + 2z cos α\| is less than 2

Answer» Correct Answer - Option 1 : less than 2

Concept:

Triangle inequality:

| z1 + z₂ | \(\leq\)| z₁| + | Z₂|

| cos α | \(\leq\) 1

Calculations:

Given, | z2 + 2z cos α | \(\leq\)| z2 | + | 2z cos α |

We know that, | cos α | \(\leq\) 1

⇒ | z2 + 2z cos α | \(\leq\)| z2 | + | 2z |

Put the value of |z| as √3 - 1

⇒ | z2 + 2z cos α | < (√3 - 1)²+ 2 (√3 - 1)

⇒ | z2 + 2z cos α | < 2

Hence, If |z| < √3 - 1, then |z2 + 2z cos α| is less than 2

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If |z|

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