If #(a,b)# is a are the coordinates of a point in Cartesian Plane, #u# is its magnitude and #alpha# is its angle then #(a,b)# in Polar Form is written as #(u,alpha)#.
Magnitude of a cartesian coordinates #(a,b)# is given by#sqrt(a^2+b^2)# and its angle is given by #tan^-1(b/a)#
Let #r# be the magnitude of #(3sqrt3,-3)# and #theta# be its angle.
Magnitude of #(3sqrt3,-3)=sqrt((3sqrt3)^2+(-3)^2)=sqrt(27+9)=sqrt36=6=r#
Angle of #(3sqrt3,-3)=Tan^-1((-3)/(3sqrt3))=Tan^-1(-1/sqrt3)=-pi/6#
#implies# Angle of #(3sqrt3,-3)=-pi/6#
This is the angle in clockwise direction.
But since the point is in fourth quadrant so we have to add #2pi# which will give us the angle in anti-clockwise direction.
#implies# Angle of #(3sqrt3,-3)=-pi/6+2pi=(-pi+12pi)/6=(11pi)/6#
#implies# Angle of #(3sqrt3,-3)=(11pi)/6=theta#
#implies (3sqrt3,-3)=(r,theta)=(6,(11pi)/6)#
#implies (3sqrt3,-3)=(6,(11pi)/6)#
Note that the angle is given in radian measure.
Also the answer #(3sqrt3,-3)=(6,-pi/6)# is also correct.
