
theorem Th6:
for k being odd Element of NAT
for x being Element of F_Complex st Re(x) = 0
holds Re((power F_Complex).(x,k)) = 0
proof
let k be odd Element of NAT;
let x be Element of F_Complex;
assume A1: Re(x) = 0;
defpred P[Nat] means
   for k1 being Element of NAT st k1 = $1 & k1 is odd
   for x being Element of F_Complex st Re(x) = 0
   holds Re((power F_Complex).(x,k1)) = 0;
A2: now let k be Nat;
   assume A3: for n being Nat st n < k holds P[n];
   now per cases by NAT_1:23;
   case A4: k = 1;
     now let k1 be Element of NAT;
       assume A5: k1 = k & k1 is odd;
       let x be Element of F_Complex;
       assume A6: Re(x) = 0;
       reconsider z = 0 as Element of NAT by ORDINAL1:def 12;
       1 = 0 + 1; then
       (power F_Complex).(x,1)
            = (power F_Complex).(x,z) * x by GROUP_1:def 7
           .= 1_(F_Complex) * x by GROUP_1:def 7
           .= x;
       hence Re((power F_Complex).(x,k1)) = 0 by A6,A5,A4;
       end;
     hence P[k];
     end;
   case k = 0;
     hence P[k];
     end;
   case k >= 2;
     then reconsider n = k-2 as Element of NAT by INT_1:5;
     reconsider n1 = n+1 as Element of NAT;
     A7: n1 + 1 = k & n + 1 = n1;
     now let k1 be Element of NAT;
       assume A8: k1 = k & k1 is odd;
       let x be Element of F_Complex;
       assume A9: Re(x) = 0;
       A10: n is odd
           proof
           consider t being Integer such that
           A11: k = 2 * t + 1 by A8,ABIAN:1;
           n = 2 * (t - 1) + 1 by A11;
           hence thesis;
           end;
       A12: now assume n >= k;
          then (k - 2) - k >= k - k by XREAL_1:11;
          hence contradiction;
          end;
       A13: (power F_Complex).(x,k1)
          = (power F_Complex).(x,n1) * x by A8,A7,GROUP_1:def 7
         .= ((power F_Complex).(x,n) * x) * x by GROUP_1:def 7
         .= (power F_Complex).(x,n) * (x * x);
       set z1 = (power F_Complex).(x,n), z2 = x * x;
       A14: Im z2 = Re x * Im x + Re x * Im x by COMPLEX1:9
               .= 0 by A9;
       A15: Re z1 = 0 by A10,A12,A3,A9;
       thus Re((power F_Complex).(x,k1))
          = Re z1 * Re z2 - Im z1 * Im z2 by A13,COMPLEX1:9
         .= 0 by A14,A15;
       end;
     hence P[k];
     end;
   end;
   hence P[k];
   end;
for k being Nat holds P[k] from NAT_1:sch 4(A2);
hence thesis by A1;
end;
