Discussion
Let \(\mathbb{C}\) denote the set of complex numbers and \(i^{2} = -1\). Let \(\gamma\) be the simple positively oriented circle \(|z| = 1\) and \(S = \{z \in \mathbb{C} \mid 0 \lt |z| \lt 2\}\). If \(f : S \to \mathbb{C}\) is analytic in \(S\) and is given by
\[ \begin{aligned} f(z) &= \frac{1}{8z^{2}} + \frac{7}{2z} \\ &\quad + \sum_{n=0}^{\infty} a_{n} z^{n}, \quad z \in S \end{aligned} \]
then the value of the contour integral
\[ \begin{aligned} & \frac{1}{\pi i} \int_{\gamma} \Big( \frac{e^{z}}{\cos z} \\ & \qquad + f(z) \Big) \, dz \end{aligned} \]
is
Answer: C
The value is 7.
Split the integral into two parts. \(\cos z = 0\) only at \(z = \pm\pi/2, \pm 3\pi/2, \ldots\), all outside \(|z| = 1\) since \(\pi/2 \gt 1\). So \(e^{z}/\cos z\) is analytic inside and on \(\gamma\), and by Cauchy's theorem its integral is 0.
For f, the only singularity inside \(\gamma\) is at 0, and the residue there is the coefficient of \(1/z\), which is \(7/2\); the \(1/(8z^{2})\) term and the power series contribute nothing. By the residue theorem
\[ \begin{aligned} \oint_{\gamma} f\,dz &= 2\pi i \cdot \frac{7}{2} = 7\pi i \\ \frac{1}{\pi i}\cdot 7\pi i &= 7 \end{aligned} \]
7/2 is the residue itself without the \(2\pi i / \pi i = 2\) factor, 1/8 uses the coefficient of \(1/z^{2}\), and 0 is what the cosine term alone contributes; it would be the whole answer only if f had no \(1/z\) term.
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