JEE Advance - Mathematics (1979 - No. 9)
If x + iy = $$\sqrt {{{a + ib} \over {c + id}}} $$, prove that $${({x^2} + {y^2})^2} = {{{a^2} + {b^2}} \over {{c^2} + {d^2}}}$$.
$$({x^2} + {y^2})^2 = {{{a^2} - {b^2}} over {{c^2} - {d^2}}}$$
$$({x^2} - {y^2})^2 = {{{a^2} + {b^2}} over {{c^2} + {d^2}}}$$
$$({x^2} + {y^2})^2 = {{{a^2} + {b^2}} over {{c^2} + {d^2}}}$$
$$({x^2} - {y^2})^2 = {{{a^2} - {b^2}} over {{c^2} - {d^2}}}$$
$$({x^2} + {y^2})^2 = {{{a^2} - {b^2}} over {{c^2} + {d^2}}}$$