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

Question

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

Charles Autterson · Accepted Answer

#1.24# #cis# #3.09# 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)#.