If the eqn ax2+bx+c=0 (0 1, |Z2| >1.

If the eqn ax²+bx+c=0  (0 <a<b<c) has non real complex roots Z₁ and Z₂

then z_{1 &} Z_{2 must be conjugate complex number.}

and b²-4ac<0

and |Z_{1|= |}Z_2|

so i f  i will prove for one root then it is also true for other root

z₁,z_{2={-b±i√(4ac-b²)}/2a}

|z₁|=√{b² +(√(4ac-b²))²}/2a 

|z₁|=√{b²+4ac-b²}/2a

        =√(c/a) >1

so |Z_{1|= |}Z_{2| >1}

As in the previous thread we note that z₁ and z₂ are complex conjugates. Hence |z₁| = |z₂|.

Now z₁ z2 = c/a and hence |z₁ z2| = |z₁|² = |c/a|>1 since c>a>0