If (a+ib)(c+id)(e+if)(g+ih)=A+iB, then ( $a^{2}$ + $b^{2}$ )( $c^{2}$ + $d^{2}$ )( $e^{2}$ + $f^{2}$ )( $g^{2}$ + $h^{2}$ )=

A²+B²

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A²-B²

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A²

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B²

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Solution

The correct option is A
A²+B²

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

⇒ (a - ib)(c - id)(e - if)(g + ih) = A - iB ........(ii)

Multiplying (i) and (ii), we get

( $a^{2}$ + $b^{2}$ )( $c^{2}$ + $d^{2}$ )( $e^{2}$ + $f^{2}$ )( $g^{2}$ + $h^{2}$ ) = $A^{2}$ + $B^{2}$

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Similar questions

If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that

(a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B².

(a + i b) (c + i d) = A + i B

(a^{2} + b^{2}) (c^{2} + d^{2}) = A^{2} + B^{2}

a, b, c, d, p

(a + i b)^{c + i d} = p > 0

(a^{2} + b^{2})^{(c^{2} + d^{2})} =

A = [\begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix}]

a^{2} + b^{2} + c^{2} + d^{2} = 1

A^{- 1}

A = [\begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix}]

a^{2} + b^{2} + c^{2} + d^{2} = 1

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