If (a,b)(a,b) is a are the coordinates of a point in Cartesian Plane, uu is its magnitude and alphaα is its angle then (a,b)(a,b) in Polar Form is written as (u,alpha)(u,α).
Magnitude of a cartesian coordinates (a,b)(a,b) is given bysqrt(a^2+b^2)√a2+b2 and its angle is given by tan^-1(b/a)tan−1(ba)
Let rr be the magnitude of (1,-sqrt3)(1,−√3) and thetaθ be its angle.
Magnitude of (1,-sqrt3)=sqrt((1)^2+(-sqrt3)^2)=sqrt(1+3)=sqrt4=2=r(1,−√3)=√(1)2+(−√3)2=√1+3=√4=2=r
Angle of (1,-sqrt3)=Tan^-1(-sqrt3/1)=Tan^-1(-sqrt3)=-pi/3(1,−√3)=tan−1(−√31)=tan−1(−√3)=−π3
implies⇒ Angle of (1,-sqrt3)=-pi/3(1,−√3)=−π3
But since the point is in fourth quadrant so we have to add 2pi2π which will give us the angle.
implies⇒ Angle of (1,-sqrt3)=-pi/3+2pi=(-pi+6pi)/3=(5pi)/3(1,−√3)=−π3+2π=−π+6π3=5π3
implies⇒ Angle of (1,-sqrt3)=(5pi)/3=theta(1,−√3)=5π3=θ
implies (1,-sqrt3)=(r,theta)=(2,(5pi)/3)⇒(1,−√3)=(r,θ)=(2,5π3)
implies (1,-sqrt3)=(2,(5pi)/3)⇒(1,−√3)=(2,5π3)
Note that the angle is given in radian measure.
Note that the answer (1,-sqrt3)=(2,-pi/3)(1,−√3)=(2,−π3) is also correct.
