Help Please

recca02

Joined Apr 2, 2007
1,211
for addition,subtraction of complex nos in R<phi always transform to rectangular form
i;e A+jB form.
where A=Rcos(phi)
and
B=Rsin(phi)
then add imaginary parts separately and real separately.
then if u wish u can convert to polar form again by
sqrt(A^2 +b^2)=R
and tan(phi)=(B/A)

in case of calculator it migh depend on ur calculator model.
try solving in complex mode (be sure to express the nos properly)

did that help?
 

Thread Starter

Mr Walker

Joined Aug 15, 2007
6
thank you very much but can you break down the equation....its complicated ut if you solve it i'm sure i can follow your steps back and figure it from there
 

recca02

Joined Apr 2, 2007
1,211
120<0 - 120<120
=120cos0 + j.120sin0 -(120cos120 + j.120sin120)
=120-120*-0.5 + j(0-103.92)
=180 - j.103.92
180=A;
(-)103.92=B---note there is a minus sign before 103.92 this means B is to be taken with sign if it wud have been '+' in front then B wud have been +ve
R= sqrt(sq(180) + sq(103.92))=207.84
phi= tan inverse of (B/A)=tan inverse of (-0.577)=-30 degrees
thus answer = R<phi
=207.84<-30
anything else?
 

Thread Starter

Mr Walker

Joined Aug 15, 2007
6
120<0 - 120<120
=120cos0 + j.120sin0 -(120cos120 + j.120sin120)
=120-120*-0.5 + j(0-103.92)
=180 - j.103.92
180=A;
(-)103.92=B---note there is a minus sign before 103.92 this means B is to be taken with sign if it wud have been '+' in front then B wud have been +ve
R= sqrt(sq(180) + sq(103.92))=207.84
phi= tan inverse of (B/A)=tan inverse of (-0.577)=-30 degrees
thus answer = R<phi
=207.84<-30
anything else?
can you explain the steps between 1 to 2 to 3....like how you got from point one to point b
never mind i firgured it out thanks
 
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