intrestingly someone posted this q on aops
check this--
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=331801
All the roots of the eqn a_1z^3+a_2z^2+a_3z+a_4=3 where |ai|<1 {i=1,2,3,4} lie outside the circle with centre at origin and radius equal to
a) 1
b) 1/3
c) 2/3
d) N.O.T.
Modelled after prophet sir's 'region of roots'.
a_1+\frac{a_2}{z}+\frac{a_3}{z^2}+\frac{a_4}{z^3}=\frac{3}{z^3}
Triangle inequality gives |z|^3+|z|^2+|z|-2>0\Rightarrow |z|>1
Thus a).
well the answer given was (c) only i wasn't satisfied with the soln given....
@Soumik : can you throw a bit more light ... maybe the link to where you saw that model of sir's
intrestingly someone posted this q on aops
check this--
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=331801
I was sleeping when I wrote the above stuff......
What I wanted to say was f(x)=x3+x2+x-2>0 for all x satisfying the eqn....x=|z|.
But f(x)<0 for b) and c). Thus ans =(b,c).