(1) Let 
where F and 
are reduced and
. Then by the formula proved above, there exists 
such that 
. As before, we deduce from the
inequalities |Q|<P<R and 
that 
, and hence
that M is an automorphism of 
. The proof then terminates as in the real
case, except that there are no special cases to consider since the forms are
irreducible.
(2) As in the real case, a complex reducible form F belonging to U must
be equivalent to 
or to 
with
. If 
and 
with
, then 
for some 
.
Hence the reduced form 
is equivalent to a multiple of 
with
or 
respectively which are also reduced,
and hence it is equal to that multiple, hence we have either Q=0 or
Q=P for the form F, which are both excluded from the definition of a
reduced form in the complex case.
(3) is clear: we can always reduce first 
by an element of 
so that it satisfies 0<|Q|<P<R, the strict inequalities being guaranteed by
the irreducibility of F. We must have 
otherwise F is reducible,
and since P=a, we have a>0. Changing 
into
if necessary (which changes Q into -Q and leave P and R
unchanged), we may also assume that 
. Finally, if b=0,
changing again 
into 
of necessary we may assume d>0
since 
.
