What is the trigonometric form of # (12-2i) #?
1 Answer
Mar 11, 2016
Explanation:
Using the following formulae :
#• r^2 = x^2 + y^2#
#• theta = tan^-1 (y/x)# here x = 12 and y = - 2
hence
#r^2 = 12^2 + (-2)^2 = 148 rArr r = sqrt148 = 2sqrt37# and
#theta = tan^-1(-2/12) ≈ -0.165" radians " #
#rArr (12-2i) = 2sqrt37 [cos(-0.165) + isin(-0.165) ]#
