How do you convert #( 1 , - sqrt3 )#into polar coordinates?

1 Answer
Jan 9, 2016

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 #(1,-sqrt3)# and #theta# be its angle.
Magnitude of #(1,-sqrt3)=sqrt((1)^2+(-sqrt3)^2)=sqrt(1+3)=sqrt4=2=r#
Angle of #(1,-sqrt3)=Tan^-1(-sqrt3/1)=Tan^-1(-sqrt3)=-pi/3#

#implies# Angle of #(1,-sqrt3)=-pi/3#

But since the point is in fourth quadrant so we have to add #2pi# which will give us the angle.

#implies# Angle of #(1,-sqrt3)=-pi/3+2pi=(-pi+6pi)/3=(5pi)/3#

#implies# Angle of #(1,-sqrt3)=(5pi)/3=theta#

#implies (1,-sqrt3)=(r,theta)=(2,(5pi)/3)#
#implies (1,-sqrt3)=(2,(5pi)/3)#
Note that the angle is given in radian measure.

Note that the answer #(1,-sqrt3)=(2,-pi/3)# is also correct.