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Algebra of Complex Numbers

If a+i bc+i d...

Question

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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A⁴ + B⁴

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A⁴ - B⁴

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Solution

The correct option is B
A² + B²

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

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

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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) (e + i f) (g + i h) = x + i y

(a^{2} + b^{2}) (c^{2} + d^{2}) (e^{2} + f^{2}) (g^{2} + h^{2}) = x^{2} + y^{2}

(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}

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