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X*4+ax*3++bx*2+cx+d be a polynomial equation with real cofficients and real zeroes.if |f(i)|=1.find a+b+c+d.i=root of -1
f(i)=1-ai-b-ci+d=11-b-d=real part=1-i(a+c)=complex part =0b+d=0a+c=0a+b+c+d=0sher mohammadb.tech, iit delhi
If p,q,r,s are the real roots of the polynomial, then f(i) = (i-p)(i-q)(i-r)(i-s) and hence |f(i)| = (1+p^2)(1+q^2)(1+r^2)(1+s^2) >=1. But |f(i)| = 1 and hence we must have p=q=r=s=0 i.e. a=b=c=d=0=a+b+c+d
@sher md. its mod of f(i).
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