If ( a + ib ) ( c + id ) ( e + if ) ( g + ih ) = A + i B, then show that ( a 2 + b 2 ) ( c 2 + d 2 ) ( e 2 + f 2 ) ( g 2 + h 2 ) = A 2 + B 2 .

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Solution

The given expression is,

( a+ib )( c+id )( e+if )( g+ih )=A+iB

By taking modulus on both the sides, we get

| ( a+ib )( c+id )( e+if )( g+ih ) |=| A+iB | [ ∵| a×b |=| a |×| b | ] | ( a+ib ) |×| ( c+id ) |×| ( e+if ) |×| ( g+ih ) |=| A+iB |

Simplify the above expression,

a 2 + b 2 × c 2 + d 2 × e 2 + f 2 × g 2 + h 2 = A 2 + B 2

By squaring both sides, we get

( a 2 + b 2 )( c 2 + d 2 )( e 2 + f 2 )( g 2 + h 2 )= A 2 + B 2

Hence, the expression has been proved.

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