Question 383150
Geometric progression means : z2=a*z1, z3=a^2*z1
 
then 
 
 sum : {{{z1*(a^3-1)/(a-1)=30}}}
 sum of square :  {{{z1^2+z1^2*a^2+z1^2*a^4=z1^2*((a^2)^3-1)/(a^2-1)=60i}}}
 
the first sum squared is : {{{z1^2*(a^3-1)^2/((a-1)^2)=900}}}
 divided by the sum of squared : 
 
{{{(a^3-1)^2/(a-1)^2*(a^2-1)/(a^6-1)=-20/3*i}}}

{{{(a^3-1)^2/(a-1)^2*((a+1)(a-1))/((a^3-1)(a^3+1))=-20/3*i}}}
 
{{{(a^3-1)/(a-1)*(a+1)/(a^3+1)=-20/3*i}}}| use (a^3+1)=(a+1)(a^2-a+1)
 
{{{(a^2-a+1)/(a^2+a+1)=-20/3*i}}}

{{{a^2-a+1=-20/3*i*(a^2+a+1)}}}

{{{(1+20/3*i)a^2+(-1+20/3*i)a+(1+20/3*i)=0}}}
 
Let {{{D =(-1+20/3*i)^2-4*(1+20/3*i)^2}}}
 
then {{{a=(-(-1+20/3*i)+-sqrt(D))/(2(1+20/3*i))}}}

and {{{z1=300/(a^2+a+1)}}}