let z be Complex; ( Arg z in ].0,(PI / 2).[ iff ( Re z > 0 & Im z > 0 ) )
A1:
Arg z < 2 * PI
by Th34;
thus
( Arg z in ].0,(PI / 2).[ implies ( Re z > 0 & Im z > 0 ) )
( Re z > 0 & Im z > 0 implies Arg z in ].0,(PI / 2).[ )
assume that
A5:
Re z > 0
and
A6:
Im z > 0
; Arg z in ].0,(PI / 2).[
z = (Re z) + ((Im z) * <i>)
by COMPLEX1:13;
then
z <> 0 + (0 * <i>)
by A5, COMPLEX1:77;
then A7:
( |.z.| > 0 & z = (|.z.| * (cos (Arg z))) + ((|.z.| * (sin (Arg z))) * <i>) )
by Def1, COMPLEX1:47;
then
sin (Arg z) > 0
by A6, COMPLEX1:12;
then A8:
sin . (Arg z) > 0
by SIN_COS:def 17;
cos (Arg z) > 0
by A5, A7, COMPLEX1:12;
then
cos . (Arg z) > 0
by SIN_COS:def 19;
then A9:
not Arg z in [.(PI / 2),((3 / 2) * PI).]
by Th14;
0 <= Arg z
by Th34;
then A10:
Arg z > 0
by A8, SIN_COS:30;
not Arg z in [.PI,(2 * PI).]
by A8, Th10;
then
( PI / 2 > Arg z or ( PI > Arg z & (3 / 2) * PI < Arg z ) )
by A1, A9, XXREAL_1:1;
hence
Arg z in ].0,(PI / 2).[
by A10, Lm5, XXREAL_0:2, XXREAL_1:4; verum