theorem Th93:
  |.z.| = 0 implies z = 0c n
proof
  assume
A1: |.z.| = 0;
  now
    let j be Nat;
    assume
A2: j in Seg n;
    then reconsider c = z.j as Element of COMPLEX by Th57;
    0 <= Sum sqr abs z by RVSUM_1:86;
    then (abs z).j = (n|->0).j by A1,RVSUM_1:91,SQUARE_1:24;
    then |.c.| = (n|-> 0).j by A2,Th88;
    then c = 0c by COMPLEX1:45;
    hence z.j = (n|->0c).j;
  end;
  hence thesis by FINSEQ_2:119;
end;
