Usually I always simplify this kind of fraction by using the formula 1/z = (zbar(z))/abs(z)^21z=z¯z|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)|−i−8|=√64+1=√65 and abs(-i + 7) = sqrt(50)|−i+7|=√50. Hence the following results : -i - 8 = sqrt(65)(-8/sqrt(65) - i/sqrt(65))−i−8=√65(−8√65−i√65) and -i + 7 = sqrt(50)(7/sqrt(50) - i/sqrt(50))−i+7=√50(7√50−i√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).
