1. The problem statement, all variables and given/known data Find and classify the singularity. f(z) = sin(z) / z 2. Relevant equations sin(z) = (z - z3 / 3! + z5 / 5! - z7 / 7! + z9 / 9! - ... 3. The attempt at a solution so the singularity is at z = 0. f(z) = (1/z) * (z - z3 / 3! + z5 / 5! - z7 / 7! + z9 / 9! - ...) f(z) = 1 - z2 / 3! + z4 / 5! - z6 / 7! + z8 / 9! ... and these are terms starting at index 0, because the series for sin(z) is Σ(-1)k * (z2k + 1 / (2k+1)!), for k >= 0. The definition of removable singularity is if cn = 0 for n = -1, -2, -3 ... then f(z) has a removable singularity at α. ANSWER: So the singularity z = 0 is removable because the series for sin(z) starts at k = 0, and so any term k<0 must have coefficient 0. i'm really confused how to check whether the singuarlity is removable, a pole, or essential. The definition doesn't make sense to me. How can I check if ck = 0 for all k < 0, if there are no terms for k < 0 in the power series for sin(z)?