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  • Answer please! - Complex numbers and quadratic equations - JEE Main-11

If \left |Z-\frac{4}{Z} \right |= 2, then the maximum value of \left | Z \right | is equal to

  • Option 1)

    \sqrt{5}+1

  • Option 2)

    2

  • Option 3)

    2+\sqrt{2}

  • Option 4)

    \sqrt{3}+1

 

Answers (2)

best_answer

As we learnt in

Triangle Law of Inequality in Complex Numbers -

\left | z_{1}+z_{2} \right |\leq \left | z_{1} \right |+\left | z_{2} \right |

- wherein

|.| denotes modulus of complex number.

 

 \left | z-\frac{4}{z}\right |=2

|z|-\frac{4}{|z|}=2

|z|^{2}-2|z|-4=0

\therefore\ \; |z|=\frac{2+ \sqrt{4+16}}{2}=\frac{2+ \sqrt{20}}{2}=1+ \sqrt{5}

Correct option is 1.

 


Option 1)

\sqrt{5}+1

This is the correct option.

Option 2)

2

This is an incorrect option.

Option 3)

2+\sqrt{2}

This is an incorrect option.

Option 4)

\sqrt{3}+1

This is an incorrect option.

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