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[Post New]17 Apr 2008 12:21:32 IST
    Subject: maths discussion
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satyabharadwaja (22)

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if  a+ib =1/(c+id) ,then the value of
(a2+b2) x(c2+ d2)=
satyabharadwaja
    
[Post New]17 Apr 2008 12:34:12 IST
    Subject: maths discussion
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Decoder (331)

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1...

take conjugate both sides ...
multiply this eq. with the previous one..
u get ans as 1..
Diamonds r formed under greatest pressures..
so r the champs.

Kriteesh..


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[Post New]17 Apr 2008 13:30:13 IST
    Subject: Re:maths discussion
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hsbhatt (3071)

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More succintly,
 
(a+ib)(c+id) = 1
 
Hence |(a+ib)(c+id)|^2 = |a+ib|^2 |c+id|^2 = (a^2+b^2)(c^2+d^2) = 1
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[Post New]17 Apr 2008 13:31:42 IST
    Subject: Re:maths discussion
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hsbhatt (3071)

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As a related exercise can you use the above result to prove that
 
(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2
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[Post New]17 Apr 2008 13:45:23 IST
    Subject: Re:maths discussion
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sboosy (2970)

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\mbox{Instead of taking modulus individually,we multiply} \\ \\ |(a+ib)(c+id)|^2 = |(ac-bd)+i(ad+bc)|^2 = (ac-bd)^2+(ad+bc)^2 \\ \\ \mbox{Thus} \ (a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2
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[Post New]17 Apr 2008 14:46:05 IST
    Subject: Re:maths discussion
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sriram.a (208)

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    (a+ib)(c+id)=1       ----------------(1)
 
(a-ib)(c-id)=1      -------------(2) 
 
multiply (1)&(2)
 
(a^2+b^2)(c^2+d^2)=1
<SRIRAM.A> on high way of IIT
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