Discussion
For a simple pole at \( z = a \), the residue of \( f \) is
Answer: D
Multiplying by the vanishing factor cancels the pole and leaves the coefficient \( a_{-1} \):
\[ \mathrm{Res}_{z=a} f \;=\; \lim_{z \to a} (z - a)f(z). \]
The square would be needed for a double pole, together with one differentiation, and the plain limit of \( f \) does not exist at a pole at all
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