Discussion
Consider the function \(f(z) = \frac{1}{z^{2}(z-2)^{3}}\) of a complex variable \(z\). The residues of the function at \(z = 0\) and \(z = 2\), respectively, are
Answer: C
The residues are \(-\frac{3}{16}\) and \(\frac{3}{16}\).
At \(z = 0\) (a pole of order 2), the residue is the first derivative of \(\frac{1}{(z-2)^{3}}\) at 0. At \(z = 2\) (order 3), it is half the second derivative of \(\frac{1}{z^{2}}\) at 2:
\[ \begin{aligned} \text{Res}_{0} &= \left.\frac{-3}{(z-2)^{4}}\right|_{0} = -\frac{3}{16} \\ \text{Res}_{2} &= \frac{1}{2}\left.\frac{6}{z^{4}}\right|_{2} = \frac{3}{16} \end{aligned} \]
Check: the function falls off as \(z^{-5}\), so the residues must sum to zero, as they do. The options with \(\frac{3}{8}\) forget the factor \(\frac{1}{2!}\) or the power of 2 in the denominator.
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