Question

No spams please. Please answer this fast.

Answer 1

Step-by-step explanation:

(i)given (a+ib)(c+id)=x+iy.......(1)

conjugate of eqn (1)

(a-ib)(c-id)=(x-iy)

(ii)(a+ib)(c+id)=x+iy.......(1)

conjugate of eqn (1)

(a-ib)(c-id)=(x-iy)......(2)

Now,(1)×(2)

(a+ib)(c+id)(a-ib)(c-id)=(x+iy)(x-iy)

=(ac+iad+ibc-bd)(ac-iad-ibc-bd)=x²+y²

={(ac-bd)+i (ad+bc)} {(ac-bd)-i(ad+bd)=

x²+y²

=(ac-bd)²-{i (ad+bc)}²=x²+y²

=(ac-bd)²+(ad+bc)²=x²+y²

thankyou

Answer 2

$(a + ib)(c + id) = x + iy \\ ac + ibc + iad + ( {i}^{2})bd = x + iy \\ ac - bd + i(bc + ad) = x + iy \\ \\ by \: comparing \: we \: get \\ \\ x = ac - bd.....eq(1) \\ y = bc + ad ......eq(2)\\ \\ 1. \\ \\ (a - ib)(c - id) = x - iy \\ \\ put \: the \: upper \: value \: here... \\ \\ (a - bi)(c - id) = ac - bd - i(bc + ad) \\ \\ ac - ibc - iad + ({i})^{2} bd = ac - bd - i(bc + ad) \\ \\ ac - bd - i(bc + ad) = ac - bd - i(bc + ad) \\ lhs = rhs..... |hence \: proved| \\ \\ (2) \\ \\ from \: eq(1) \: and \:eq (2) \\ \\ x + iy = (ac - bd) + i(bc - ad) \\ \\ |x + iy| = \sqrt{ {(ac - bd)}^{2} + {(bc - ad)}^{2} } \\ \\ { |x + iy| }^{2} = {(ac - bd)}^{2} + {(bc - ad)}^{2} \\ \\ |x + iy| = \sqrt{ {x}^{2} + {y}^{2} } .....from \: here \\ we \: get \\ \\ {x}^{2} + {y}^{2} = {(ac - bd)}^{2} + {(bc - ad)}^{2} \\ \\ |hence \: proved|$

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