MCQ
If $x + iy = \sqrt {\frac{{a + ib}}{{c + id}}} ,$then ${({x^2} + {y^2})^2} = $
  • $\frac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}$
  • B
    $\frac{{a + b}}{{c + d}}$
  • C
    $\frac{{{c^2} + {d^2}}}{{{a^2} + {b^2}}}$
  • D
    ${\left( {\frac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}} \right)^2}$

Answer

Correct option: A.
$\frac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}$
a
(a) $x + iy = \sqrt {\frac{{a + ib}}{{c + id}}} $==> $x - iy = \sqrt {\frac{{a - ib}}{{c - id}}} $
Also ${x^2} + {y^2} = (x + iy)(x - iy) = \sqrt {\frac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}} $
==> ${({x^2} + {y^2})^2} = \frac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}$

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