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Question

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

Show that,
i) aibcid=xiy
ii)a2+b2c2+d2=x2+y2

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Solution

Consider the given that,

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

Part (1)

prove that, (aib)(cid)=xiy

Proof:- By equation (1),

Take conjugate and we get,

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

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

aibcid=xiyHenceproved.......(2)


Part (2),

Multiplying (1) and (2) to and we get,

(a+ib)(c+id)×(aib)(cid)=(x+iy)(xiy)

a2i2b2c2i2d2=x2i2y2

a2+b2c2+d2=x2+y2Henceproved.


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