Discussion
The residue of \(f(z)\) at its pole is equal to 1. Then the value of \(\alpha\) is
Answer: D
The value is \(\alpha = 3\).
\(f(z) = \frac{z^{2} + \alpha z}{(z + 1)^{2}}\) has a double pole at \(z = -1\) (the numerator is \(1 - \alpha \ne 0\) there unless \(\alpha = 1\)). For a pole of order 2, the residue is the derivative of \((z + 1)^{2}f(z)\) at the pole:
\[ \begin{aligned} \operatorname{Res}_{z=-1} f &= \frac{d}{dz}(z^{2} + \alpha z)\Big|_{z=-1} \\ &= -2 + \alpha = 1 \Rightarrow \alpha = 3 \end{aligned} \]
\(\alpha = 1\) would cancel a factor \((z + 1)\) and leave a simple pole with residue -1; \(\alpha = -1\) gives residue -3 and \(\alpha = 2\) gives 0. Only \(\alpha = 3\) gives residue 1.
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