What is the trigonometric form of # (-6-i) #?
1 Answer
Mar 8, 2016
Explanation:
Using the following formulae:
#• r^2 = x^2 + y^2 #
#• theta = tan^-1(y/x) # here x = -6 and y = -1
#r^2 = (-6)^2+(-1)^2 = 37 rArr r = sqrt37# and
# theta = tan^-1((-1)/(-6)) = tan^-1(1/6) ≈0.165 " radians "#
#rArr (-6-i) = sqrt37(cos(0.165) + isin(0.165)) #
