Precalculus, Chapter 6, 6.5, Section 6.5, Problem 16

z=5/2(sqrt3-i)
z=(5sqrt3)/2-5/2i
r=sqrt[((5sqrt3)/2)^2+(-5/2)^2]=sqrt[75/4+25/4]=sqrt25=5
theta=arctan[(-5/2)/((5sqrt3)/2)]=arctan(-1/sqrt3)=(11pi)/6
z=5[cos((11pi)/6)+isin((11pi)/6)]