Since we are given the Cartesian coordinates, we are given #x# and #y# . For polar coordinates, we need to figure out #r# and #theta# . #r# is easy, we just use Pythagorean:
To figure out #theta# , I like to use cosine because the range of arccosine is in quadrants I and II and adjusting #theta'# is easier. So,
#theta'=cos^(-1)x/r#
If #y>=0# then #theta=theta'# .
If #y
#r=sqrt((-3)^2+(3sqrt3)^2)=sqrt(36)=6#
#theta'=cos^(-1)((-3)/6)=(2pi)/3#
#y So the polar coordinates are #(6, (4pi)/3)# .
Judy O. · · Aug 19 2014
I presume we're looking for a radius #r# and angle #theta# such that #a + bi = r(cos theta + i sin theta)# . Pythagoras theorem gives us #r = sqrt(a^2+b^2)# . Simple trigonometry gives us #tan theta = b/a# , so #theta = arctan (b/a)# .
George C. · 1 · May 10 2015
