How do you divide # (4-8i)/(9+i) # in trigonometric form?

1 Answer
Feb 13, 2016

Answer:

#R = .337 + -915i #

Explanation:

#R = C_1/C_2," where " C_1 = 4 -8i" and " C_2 = 9 + i#
Convert each complex number to its polar form:
#C_1 = |C_1|/_theta_1 and " C_2 = |C_2| |/_theta_2#
Then, #R = |C_1|/|C_2| /_theta_1-theta_2 #
#C_1=sqrt(4^2 + (-8)^2)=sqrt(80); C_2=sqrt(9^2 + (1)^2)=sqrt(82) #
#theta_1 = tan^-1 (-2); theta_2 = tan^-1 (1/9); #
#theta_R=theta_1-theta_2 =tan^-1 (-2) - tan^-1 (1/9) = -1.218 " rads"#
Thus #R = .975 /_(theta_R = -1.218)# this is now in the polar form
#R=r/_theta_R; r = .975 " the magnitude of " R, theta_R=-69.775^o#
Now you can convert back to the rectangular coordinate
#R = r_x + r_y; R_x = |r|costheta_R + |r|sintheta_R#
#r_x = .975cos69.8 ~~.337; r_y=.975sin(-69.8) ~~-.915 #
#R ~~.337 + -915i #