Question

if x +¡y=root a+¡b/c+¡d prove that x ^2+y^2=a ^2+b ^2/c^2+d​

Answer 1

Step-by-step explanation:

If a+ibc+id=x+iy, prove that(i) a−ibc−id=(x−iy) (ii) a2+b2c2+d2=(x2+y2)

Answer 2

Step-by-step explanation:

We have,

$x + iy = \sqrt{ \frac{a + ib}{c + id} }$

$= > (x + iy)^{2} = \frac{(a + ib)(c - id)}{c^{2} + d^{2} }$

$= > ({x}^{2} - {y}^{2} ) + 2xyi = \frac{(ac + bd) + (bc - ad)i}{c ^{2} + d^{2} }$

Taking modulus and squaring it,

$= > |({x}^{2} - {y}^{2} ) + 2xyi | ^{2} = |\frac{(ac + bd) + (bc - ad)i}{c ^{2} + d^{2} } |^{2}$

$= > ({x}^{2} - {y}^{2} )^{2} + 4x^{2} y^{2} = \frac{(ac + bd) ^{2} + (bc - ad)^{2} }{(c ^{2} + d^{2} )^{2} }$

$= > ( {x}^{2} + {y}^{2} )^{2} = \frac{ {a}^{2} {c}^{2} + {b}^{2} {d}^{2} + 2abcd + {b}^{2} {c}^{2} + {a}^{2} {d}^{2} - 2abcd }{( {c}^{2} + {d}^{2} ) ^{2} }$

$= > ( {x}^{2} + {y}^{2})^{2} = \frac{ {c}^{2}( {a}^{2} + {b}^{2} ) + {d}^{2}( {a}^{2} + {b}^{2} )}{( {c}^{2} + {d}^{2})^{2} }$

$= > ( {x}^{2} + {y}^{2} )^{2} = \frac{ {a}^{2} + {b}^{2} }{ {c}^{2} + {d}^{2} }$