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^

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4-1.Complex numbers

medium

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

${A^2} + {B^2}$

${A^2} - {B^2}$

${A^2}$

${B^2}$

Solution

(a)$(a + ib)(c + id)(e + if)(g + ih) = A + iB$…..$(i)$
==> $(a – ib)(c – id)(e – if)(g – ih) = A – iB$……$(ii)$
Multiplying $(i) $ and $(ii)$, we get
$({a^2} + {b^2})({c^2} + {d^2})({e^2} + {f^2})({g^2} + {h^2}) = {A^2} + {B^2}$

Standard 11

Mathematics

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medium

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

Solution

Similar Questions

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