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

$(a + i b) (c + i d) = A + i B \to (1)$
$Replace i by - i,$
$A = iB = (a - i b) (c - i d) \to (2)$
$(A + i B) (A - i B) = (a + i b) (c + i d) (a - i b) (c - i d)$
$= (a - ib) (a + i b) (c + i d) (c - i d)$
$= (a^{2} - {(i b)}^{2}) (c^{2} - {(i d)}^{2})$
$= (a^{2} + b^{2}) (c^{2} + d^{2})$
$(∵ i^{2} = - 1)$

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