Discussion
Let \(\nu : [0, 1] \to \mathbf{C}\) be defined by \(\nu(t) = e^{4\pi it}\). Then :
\[ \frac{1}{2\pi i}\int_{\nu} \frac{e^{z}}{z^{3}}\,dz = \]
Answer: B
The value is 1.
The curve \(\nu(t) = e^{4\pi it}\), \(0 \le t \le 1\), runs round the unit circle twice, so its winding number about 0 is 2. By the residue theorem,
\[ \begin{aligned} &\frac{1}{2\pi i}\int_{\nu} \frac{e^{z}}{z^{3}}\,dz \\ &= n(\nu, 0)\,\mathrm{Res}_{z=0}\frac{e^{z}}{z^{3}} \end{aligned} \]
The residue is the coefficient of \(z^{2}\) in \(e^{z}\), which is \(1/2\). So the value is \(2 \cdot \tfrac{1}{2} = 1\).
Forgetting the double winding gives 1/2 (not offered); 2 is the winding number alone; and \(4\pi i\) is the integral without the factor \(1/(2\pi i)\) and with residue 1.
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