If x - iy = a - ib c - id prove that ( x^2 + y^2 )^2 = a^2 + b^2 c^2 + d^2

Here x - iy = a - ib c - id Squaring both sides, we get (x - iy)^2 = a - ib c - id | (x - iy) ^2 | = | a - ib c - id | | (x - iy) | | x - iy | = | a - ib c - id | ( x^2 + y^2 ) ( x^2 + y^2 ) = a^2 + b^2 c^2 + d^2 ( x^2 + y^2 ) = a^2 + b^2 c^2 + d^2 Squaring both sides ( x^2 + y^2 )^2 = a^2 + b^2 c^2 + d^2

If x - iy = a - ib c - id prove that ( x^2 + y^2 )^2 = a^2 + b^2 …

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Question

If $x - iy = \sqrt {\frac{{a - ib}}{{c - id}}} $ prove that ${({x^2} + {y^2})^2} = \frac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}$

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Answer

Here $x - iy = \sqrt {\frac{{a - ib}}{{c - id}}} $
Squaring both sides, we get
${(x - iy)^2} = \frac{{a - ib}}{{c - id}}$
$ \Rightarrow \left| {{{(x - iy)}^2}} \right| = \left| {\frac{{a - ib}}{{c - id}}} \right|$$ \Rightarrow \left| {(x - iy)} \right|\left| {x - iy} \right| = \left| {\frac{{a - ib}}{{c - id}}} \right|$
$ \Rightarrow \left( {\sqrt {{x^2} + {y^2}} } \right)\left( {\sqrt {{x^2} + {y^2}} } \right)$$ = \frac{{\sqrt {{a^2} + {b^2}} }}{{\sqrt {{c^2} + {d^2}} }} \Rightarrow ({x^2} + {y^2}) = \sqrt {\frac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}} $
Squaring both sides
${({x^2} + {y^2})^2} = \frac{{{a^2} + {b^2}}}{{{c^2} + {d^2}}}$