If $(a + i b) (c + i d) = A + i B$ , then show that $(a^{2} + b^{2}) (c^{2} + d^{2}) = A^{2} + B^{2}$ .

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Solution

$(a + i b) (c + i d) = A + i B (1)$

$(a - i b) (c - i d) = A - i B (2)$

$M u l t i p l y .$

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

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

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

\frac{(a + i b)}{(c + i d)} = x + i y

\frac{a - i b}{c - i d} = x - i y

\frac{a^{2} + b^{2}}{c^{2} + d^{2}} = x^{2} + y^{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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