
theorem Th5:
for k being even Element of NAT
for x being Element of F_Complex st Re(x) = 0
holds Im((power F_Complex).(x,k)) = 0
proof
let k be even 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 even
   for x being Element of F_Complex st Re(x) = 0
   holds Im((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 = 0;
     now let k1 be Element of NAT;
       assume A5: k1 = k & k1 is even;
       let x be Element of F_Complex;
       assume Re(x) = 0;
       (power F_Complex).(x,0) = 1_(F_Complex) by GROUP_1:def 7
                              .= 1r by COMPLFLD:def 1;
       hence Im((power F_Complex).(x,k1)) = 0 by A4,A5,COMPLEX1:6;
       end;
     hence P[k];
     end;
   case k = 1;
     hence P[k] by Lm1;
     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;
     A6: n1 + 1 = k & n + 1 = n1;
     now let k1 be Element of NAT;
       assume A7: k1 = k & k1 is even;
       let x be Element of F_Complex;
       assume A8: Re(x) = 0;
       A9: n is even
           proof
           consider t being Nat such that
           A10: k = 2 * t by A7,ABIAN:def 2;
           n = 2 * (t - 1) by A10;
           hence thesis;
           end;
       A11: now assume n >= k;
          then (k - 2) - k >= k - k by XREAL_1:11;
          hence contradiction;
          end;
       A12: (power F_Complex).(x,k1)
          = (power F_Complex).(x,n1) * x by A7,A6,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;
       A13: Im z2 = Re x * Im x + Re x * Im x by COMPLEX1:9
               .= 0 by A8;
       A14: Im z1 = 0 by A9,A11,A3,A8;
       thus Im((power F_Complex).(x,k1))
          = Re z1 * Im z2 + Re z2 * Im z1 by A12,COMPLEX1:9
         .= 0 by A13,A14;
       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;
