let z be complex number ; ( Arg z in ].PI ,((3 / 2) * PI ).[ iff ( Re z < 0 & Im z < 0 ) )
thus
( Arg z in ].PI ,((3 / 2) * PI ).[ implies ( Re z < 0 & Im z < 0 ) )
( Re z < 0 & Im z < 0 implies Arg z in ].PI ,((3 / 2) * PI ).[ )proof
assume A1:
Arg z in ].PI ,((3 / 2) * PI ).[
;
( Re z < 0 & Im z < 0 )
then
PI < Arg z
by XXREAL_1:4;
then A2:
PI / 2
< Arg z
by Lm2, XXREAL_0:2;
A3:
Arg z > PI
by A1, XXREAL_1:4;
then
z <> 0
by Def1;
then A4:
(
z = (|.z.| * (cos (Arg z))) + ((|.z.| * (sin (Arg z))) * <i> ) &
|.z.| > 0 )
by Def1, COMPLEX1:133;
Arg z < (3 / 2) * PI
by A1, XXREAL_1:4;
then
Arg z in ].(PI / 2),((3 / 2) * PI ).[
by A2, XXREAL_1:4;
then
cos . (Arg z) < 0
by Th29;
then
cos (Arg z) < 0
by SIN_COS:def 23;
hence
Re z < 0
by A4, COMPLEX1:28;
Im z < 0
Arg z < (3 / 2) * PI
by A1, XXREAL_1:4;
then
Arg z < 2
* PI
by Lm6, XXREAL_0:2;
then
Arg z in ].PI ,(2 * PI ).[
by A3, XXREAL_1:4;
then
sin . (Arg z) < 0
by Th25;
then
sin (Arg z) < 0
by SIN_COS:def 21;
hence
Im z < 0
by A4, COMPLEX1:28;
verum
end;
assume that
A5:
Re z < 0
and
A6:
Im z < 0
; Arg z in ].PI ,((3 / 2) * PI ).[
z = (Re z) + ((Im z) * <i> )
by COMPLEX1:29;
then
z <> 0 + (0 * <i> )
by A5, COMPLEX1:163;
then A7:
( |.z.| > 0 & z = (|.z.| * (cos (Arg z))) + ((|.z.| * (sin (Arg z))) * <i> ) )
by Def1, COMPLEX1:133;
then
cos (Arg z) < 0
by A5, COMPLEX1:28;
then
cos . (Arg z) < 0
by SIN_COS:def 23;
then A8:
not Arg z in [.((3 / 2) * PI ),(2 * PI ).]
by Th32;
sin (Arg z) < 0
by A6, A7, COMPLEX1:28;
then
sin . (Arg z) < 0
by SIN_COS:def 21;
then A9:
not Arg z in [.0 ,PI .]
by Th24;
Arg z < 2 * PI
by Th52;
then A10:
Arg z < (3 / 2) * PI
by A8, XXREAL_1:1;
0 <= Arg z
by Th52;
then
Arg z > PI
by A9, XXREAL_1:1;
hence
Arg z in ].PI ,((3 / 2) * PI ).[
by A10, XXREAL_1:4; verum