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Question

Consider az2+bz+c=0. where a,b,cR and 4ac>b2
If z1 and z2 are the roots of the equation given above, then which one of the following complex numbers is purely real ?

A
z1¯z2
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B
¯z1z2
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C
z1z2
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D
(z1z2)i
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Solution

The correct option is D (z1z2)i
Since, Z1 and Z2 are conjugated pairs,
let Z1=a+ib
and Z2=aib
So, Z1¯Z2=(a+ib)(aib)=a2b2+2aib
¯Z1Z2=(aib)(aib)=a2b22aib
Z1Z2=2aib
(Z1Z2)i=(2aib)i=2ab=Purely real

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