I have the following information
three phase fault : I = 6060.08 <-175.6
Single phase fault: I = 4618.63 <170.7

How do I convert this to get the rectangular form, (R + jX) ??

 

This looks like a homework problem and will get more attention over there.

 

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-2/polar-rectangular-notation/

 

Hi,

You could use the relations:
A=sqrt(R^2+I^2)
Ph=atan2(I,R)

to find R and I, which are the real and imaginary parts.
A is the amplitude of the sine, Ph is the phase.
So for:
100*sin(wt+0.2)

the amplitude A=100 and the phase Ph=0.2 in radians.

 

The best way to do it is to draw the right triangle.
You are given the hypotenuse of the right triangle, and the angle from the origin.



You have right triangle.
Hypotenuse is 4618.63
The angle inside the triangle is 180°-170.7°=9.3°
You need to find the R part and X part.

cos(9.3°)=R part/hypotenuse
cos(9.3°)=R part/4618.63
R part= 4618.63*cos(9.3°)
This gives you the magnitude of R part.
Since R part lies on the left of the origin, it will have negative sign.

You should be able to figure out the X part on your own.