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B
x2+y2
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C
(x+y)2
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D
(x−y)2
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Solution
The correct option is Bx2+y2 Given : a+ibc+id=x+iy
Now, a2+b2c2+d2=(a+ib)(a−ib)(c+id)(c−id)=a+ibc+id×a−ibc−id=a+ibc+idׯ¯¯¯¯¯¯¯¯¯¯¯¯¯a+ib¯¯¯¯¯¯¯¯¯¯¯¯¯¯c+id=(x+iy)×(x−iy)=x2+y2