If (A+iB)= \frac{\sqrt {1+2i } }{\sqrt {3+4i }}     , then (A^{2}+B^{2})^{2}    is equal to:

  • Option 1)

    5

  • Option 2)

    \frac{1}{5}

  • Option 3)

    \frac{2}{5}

  • Option 4)

    \frac{5}{2}

 

Answers (1)

As we learrnt in 

Property of Modulus of z(Complex Number) -

\left |\frac{z_{1}}{z_{2}} \right |=\frac{\left |z_{1} \right |}{\left |z_{2} \right |}

- wherein

|.| denotes modulus of complex number

A+iB =\frac{\sqrt{1+2i}}{\sqrt{3+4i}}

Taking modulus on both sides

 \left | A+iB \right |=\frac{\left | \sqrt{1+2i} \right |}{\left | \sqrt{3+4i} \right |}= > \sqrt{A^{2}+B^{2}}=\sqrt{\frac{\sqrt{5}}{\sqrt{25}}}\\*\\* Thus\: \: \: \left ( A^{2}+B^{2} \right )=\frac{1}{5}


Option 1)

5

Incorrect

Option 2)

\frac{1}{5}

Correct

Option 3)

\frac{2}{5}

Incorrect

Option 4)

\frac{5}{2}

Incorrect

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