Question

pls solve this tennetiraj86 pls I want the answer by today​

Answer 1

To ProvE :

$\tt \: \: \: \: \: \dagger{\bigg( {x}^{2} + {y}^{2} \bigg)}^{2} = \dfrac{ {a }^{2} + {b }^{2}}{{c}^{2} + {d }^{2} }$

SolutioN :

$\tt \dagger \: \: \: \: \: If \: x + iy = \sqrt{ \dfrac{a+ib}{c+id} } ,then$

$\tt prove \: that \: \Big(x^2+y^2\Big)= \dfrac{a^2+b^2}{c^2+d^2}$

$\tt : \implies x + iy = \sqrt{ \dfrac{a+ib}{c+id} }\: \: \: \: \: - ( 1 )$

✎ Taking Bar both sides.

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

✎ Now, Multiple Equation ( 1 ) and ( 2 )

$\tt : \implies x + iy \times{x - iy} = \sqrt{ \dfrac{ \overline{a+ib}}{ \overline{c+id}} } \times \sqrt{ \dfrac{ a+ib}{ c+id} }$

$\tt : \implies {x}^{2} - {i^2y}^{2} = \sqrt{ \dfrac{\overline{a+ib}}{\overline{c+id}} } \times \sqrt{ \dfrac{ a+ib}{ c+id} }$

$\tt : \implies {x}^{2} - {i^2y}^{2} = \sqrt{ \dfrac{ (a - ib)(a + ib)}{( c- id)(c +id)}}$

$\tt : \implies {x}^{2} + {y}^{2} = \sqrt{ \dfrac{ {a }^{2} - {(ib) }^{2}}{{c}^{2} -{(id) }^{2} }}$

✎ Now, Squaring Both sides We get,

$\tt : \implies{ \bigg( {x}^{2} + {y}^{2} \bigg)}^{2} = \dfrac{ {a }^{2} + {b }^{2}}{{c}^{2} + {d}^{2} }$

✡ Hence Proved.

$\rule{200}3$

MorE InformatioN :

• ( a + b )( a - b ) = a² - b².
• ( i )² = - 1.

Z = a + bi

Where as,

• Z ( Complex number )
• a ( real number )
• b ( imaginary number )

Answer 2

The answer refers to attatchment