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

A^{2} + B^{2}

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A^{2} - B^{2}

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A^{2}

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B^{2}

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Solution

The correct option is C $A^{2} + B^{2}$
Given,
$(a + i b) (c + i d) (e + i f) (g + i h) = A + i B$
Now taking modulus both sides we get,
$| (a + i b) (c + i d) (e + i f) (g + i h) | = | A + i B |$
or, $| (a + i b) | . | (c + i d) | . | (e + i f) | . | (g + i h) | = | A + i B |$ [ Since $! z_{1} . z_{2} | = | z_{1} | . | z_{2} |$ for complex number $z_{1}$ and $z_{2}$ .]
or, $\sqrt{a^{2} + b^{2}} . \sqrt{c^{2} + d^{2}} . \sqrt{e^{2} + f^{2}} . \sqrt{g^{2} + h^{2}} = \sqrt{A^{2} + B^{2}}$
Now squaring both sides we get,
$(a^{2} + b^{2}) (c^{2} + d^{2}) (e^{2} + f^{2}) (g^{2} + h^{2}) = A^{2} + B^{2}$ .

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Similar questions

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

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

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

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

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

c o s (i l o g \frac{a - i b}{a + i b})

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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