 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/4/b/solution

g

gives

\frac{g ( z )}{z - z _{0}} = \frac{g ( z _{0} )}{z - z _{0}} + g^{'} (z_{0}) + O (z - z_{0}) .

(z - z_{0})^{- 1}

, and therefore the residue, is

Res_{z = z_{0}} \frac{g ( z )}{z - z _{0}} = g (z_{0}) .

g (z_{0}) = 0

, the apparent pole is a removable singularity and the residue is zero.

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