Discussion
Residue of cot (z) at z = 0 is :
Answer: D
The residue is 1.
Write \(\cot z = \cos z / \sin z\). At \(z = 0\), \(\sin z\) has a simple zero and \(\cos 0 = 1\), so the pole is simple. For a quotient \(g/h\) with \(h\) having a simple zero, the residue is \(g(0)/h'(0)\):
\[ \text{Res}_{z=0} \cot z = \frac{\cos 0}{\cos 0} = 1 \]
The same follows from the Laurent series \(\cot z = \frac{1}{z} - \frac{z}{3} - \dots\), whose \(1/z\) coefficient is 1. The value 0 would mean no pole at all, and 1/2 or 2 come from mishandling the derivative of \(\sin z\).
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