How do you convert #-1+(sqrt3i)# to polar form?

1 Answer
Jan 14, 2017

#(r,theta)=(2,120^@)#

Explanation:

For a complex number in the form
#color(white)("XXX")a+bi#
the radius is given by the Pythagorean Theorem as
#color(white)("XXX")r=sqrt(a^2+b^2)#
In this case
#color(white)("XXX")r=sqrt((-1)^2_(sqrt(3))^2)=sqrt(1+3)=sqrt(4)=2#

The angle can be calculated using an inverse trig function and adjusting to place the point in the proper quadrant.

In this case #-1+sqrt(3)i# is in Quadrant 2 of the complex plane.

Using the #arctan# function gives an angle in #(-90^@,+90^@)#
so it will be necessary to add #180^@# to shift the angle into Q 2.
#color(white)("XXX")theta=arctan(-sqrt(3)/1)+180^@ =120^@#

(In this particular case, we have one of the standard triangles and the angle could easily be inferred with a quick sketch. )