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A function (Discussion)

koustabhdolui said – Fri, 06 Jun 2008 19:26:22 -0000 ( Link )

f(z)=az^2+bz+c

f(z) is real if z is real and f(z) is imaginary when z is imaginary.Prove that a=0

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  1. baishali said – Sun, 10 Aug 2008 07:54:51 -0000 ( Link )

    let z=1 f(1)=a2+b+c if b=c=0 then f(1)=a2 if a=1 then f(1)=1

    let z=2+3i f(2+3i)=a(2+3i)2+b(2+3i)+c now if b=c=0 then f(2+3i)=a(2+3i)2 =a(4+12i-9) =a(12i-9) therefore a=0

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