Usually I always simplify this kind of fraction by using the formula #1/z = (zbar(z))/abs(z)^2# so I'm not sure what I'm going to tell you works but this is how I'd solve the problem if I only wanted to use trigonometric form.
#abs(-i - 8) = sqrt(64+1) = sqrt(65)# and #abs(-i + 7) = sqrt(50)#. Hence the following results : #-i - 8 = sqrt(65)(-8/sqrt(65) - i/sqrt(65))# and #-i + 7 = sqrt(50)(7/sqrt(50) - i/sqrt(50))#
You can find #alpha, beta in RR# such that #cos(alpha) = -8/sqrt(65)#, #sin(alpha) = -1/sqrt65#, #cos(beta) = 7/sqrt50# and #sin(beta) = -1/sqrt50#.
So #alpha = arccos(-8/sqrt65) = arcsin(-1/sqrt65)# and #beta = arccos(-7/sqrt50) = arcsin(-1/sqrt50)#, and we can now say that #-i - 8 = sqrt(65)e^arccos(-8/sqrt65)# and #-i + 7 = sqrt(50)e^arccos(-7/sqrt50)#.
