How do you convert #sqrt(3)i - 1# to polar form?

1 Answer
Jim G.
May 23, 2016

#(2,(2pi)/3)#

Explanation:

Using the formulae that links Cartesian to Polar coordinates.

#•r=sqrt(x^2+y^2)" and " theta=tan^-1(y/x)#

now #sqrt3 i-1=-1+sqrt3 i#

here x = -1 and y =#sqrt3#

#rArrr=sqrt((-1)^2+(sqrt3)^2)=sqrt4=2#

and #theta=tan^-1(-sqrt3)=-pi/3#

#-1+sqrt3 i" is in the 2nd quadrant"#

so #theta" requires to be an angle in 2nd quadrant"#

#rArrtheta=(pi-pi/3)=(2pi)/3#

#rArr(-1,sqrt3)=(2,(2pi)/3)#

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