Vector Algebra

Question

If $open vertical bar straight Z space equals space 4 over straight z close vertical bar$ , then the maximum value of |Z| is equal to

$square root of 3 space plus 1$
$square root of 5 space plus 1$

2

$2 space plus space square root of 2$

Solution

square root of 5 space plus 1

open vertical bar straight Z minus 4 over straight z close vertical bar space equals 2 Now comma space vertical line straight z vertical line space equals space open vertical bar straight Z space minus space 4 over straight z plus 4 over straight z close vertical bar space less or equal than space 2 space plus fraction numerator 4 over denominator vertical line straight z vertical line end fraction space less or equal than 0 rightwards double arrow space 0 space less than space vertical line straight z vertical line space less or equal than space space 1 plus square root of 5

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