How do you divide # (3+4i) / (9-4i) # in trigonometric form?

1 Answer
Feb 26, 2016

#0.51(cos(1.35)+isin(1.35))#

Explanation:

Let the quotient be #q=3+4i#
and the divisor be #d=9-4i#

In trigonometric form:
#color(white)("XXX")q=r_q(cos(theta_q)+isin(theta_q))#
with
#color(white)("XXX")r_q=sqrt(3^2+4^2)=5#
and
#color(white)("XXX")theta_q= arctan(4/3)~~0.927295#

Similaryly, in trigonometric form:
#color(white)("XXX")d=r_d(cos(theta_d)+isin(theta_d))#
with
#color(white)("XXX")r_d=sqrt(9^2+(-4)^2) =sqrt(97)~~9.848858#
and
#color(white)("XXX")theta_d=arctan(9/(-4))~~-0.41822#

Using trigonometric form:
#color(white)("XXX")q/d = (r_q)/(r_d)(cos(theta_q-theta_d)+isin(theta_q-theta_d))#

#color(white)("XXX")=5/sqrt(97)(cos(0.927295-(-041822))+isin(0.927295-(-0.41822))#

#color(white)("XXX")~~0.51(cos(1.35)+isin(1.35))#