How do you divide #( -i-5) / (i +4 )# in trigonometric form?

1 Answer
Jul 13, 2017

#1.24# #cis# #3.09#

Explanation:

First, divide the expression so that you can get a complex number in #a+bi# form. To do this, you must multiply the numerator and denominator by the conjugate, #i-4#.

#(-i-5)/(i+4)#

#=(-i-5)/(i+4) * (i-4)/(i-4)# #->#multiply by the conjugate

#=(-i^2 + 4i - 5i +20)/(i^2-4^2)# #->#expand

#=(1 + 4i - 5i +20)/(-1-16)# #->#simplify

#=(21 -i)/(-17)# #->#combine like terms

#=color(blue)(-21/17 + 1/17i)# #->#rewrite in #a+bi# form

To convert this to trigonometric form, you must find out #r#, the distance from the origin to the point, and #theta#, the angle. Use the following formulas:

#r=sqrt(a^2+b^2)#

#tan theta = b/a#

In this case, #a=-21/17# and #b=1/17#.

#r=sqrt(a^2+b^2) = sqrt((-21/17)^2+(1/17)^2) ~~ color(blue)1.24#
#tan theta = (1/17)/(-21/17) = 1/17 * -17/21 = -1/21#

#theta=tan^-1(-1/21) ~~ -0.05#

However, since the coordinate #(-21/17 + 1/17i)# is in Quadrant #II#, this angle is wrong. This is because we used the #arctan# function, which doesn't account for angles outside of the range #[-pi/2, pi/2]#. To fix this, add #pi# to #theta#.

#-0.05 + pi = color(blue)3.09#

So, the trigonometric form is #1.24# #cis# #3.09#, or #1.24 (cos 3.09 + i sin 3.09)#.