Question

if given that-
$x - iy = \sqrt{ \frac{a - i}{c - id} }$
then prove that -
${x}^{2} + {y}^{2} = \sqrt{ \frac{ {a}^{2} + {b}^{2} }{ {c}^{2} + {d}^{2} } }$


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Answer 1

GIVEN :–

$\\ \implies \bf x - iy = \sqrt{ \dfrac{a - ib}{c - id}}$

TO PROVE :–

$\\ \implies \bf {x}^{2} + {y}^{2} = \sqrt{ \dfrac{ {a}^{2} + {b}^{2} }{{c}^{2} + {d}^{2}}}$

SOLUTION :–

$\\ \implies \bf x - iy = \sqrt{ \dfrac{a - ib}{c - id}} - - - eq.(1)$

• Replace i with -i :–

$\\ \implies \bf x - ( - i)y = \sqrt{ \dfrac{a -( - i)b}{c -( - i)d}}$

$\\ \implies \bf x + iy = \sqrt{ \dfrac{a +ib}{c + id}} \: \: \: - - - eq.(2)$

• Now multiply eq.(1) & eq.(2) –

$\\ \implies \bf(x - iy)(x + iy)= \sqrt{ \dfrac{a - ib}{c - id}} \sqrt{ \dfrac{a +ib}{c + id}}$

$\\ \implies \bf(x - iy)(x + iy)= \sqrt{ \dfrac{(a - ib)(a +ib)}{(c - id)(c + id)}}$

• Using identity –

$\\ \implies \bf(a-b)(a+b)= {a}^{2} - {b}^{2}$

$\\ \implies \bf{(x)}^{2} - (iy)^{2}= \sqrt{ \dfrac{(a)^{2} - (ib)^{2}}{(c)^{2} - (id)^{2}}}$

$\\ \implies \bf{(x)}^{2} - (i)^{2} (y)^{2}= \sqrt{ \dfrac{(a)^{2} - (i)^{2} (b)^{2}}{(c)^{2} - (i)^{2} (d)^{2}}}$

• We know that –

$\\ \longrightarrow \bf(i)^{2} = - 1$

$\\ \implies \bf{(x)}^{2} - ( - 1) (y)^{2}= \sqrt{ \dfrac{(a)^{2} - ( - 1) (b)^{2}}{(c)^{2} - ( - 1)(d)^{2}}}$

$\\ \implies \bf {x}^{2} + {y}^{2} = \sqrt{ \dfrac{ {a}^{2} + {b}^{2} }{{c}^{2} + {d}^{2}}}$

$\\ \large \longrightarrow{ \boxed{\bf Hence \: \: Proved }}$

Answer 2

−

so this is your answer ..

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hope it is helpful...