Discussion
The value of the contour integral
\[ \int_{\Gamma} \frac{(1-z)(3\cos z+5\sin z)}{1-z^{101}}dz, \]
where \(\Gamma = \left\{ z \in \mathbb{C} : |z| = \frac{7}{9} \right\}\), oriented in the counter-clockwise direction, is equal to
Answer: C
The integral equals 0.
The integrand can only fail to be analytic where \(1 - z^{101} = 0\), that is at the 101st roots of unity, all of which lie on the unit circle \(|z| = 1\). The contour \(|z| = \frac{7}{9}\) lies strictly inside the unit circle, so the integrand is analytic on and inside \(\Gamma\).
By Cauchy's theorem the integral is 0. The factor \(1 - z\) cancels the root at \(z = 1\), but that does not matter here since no root is enclosed. Values such as \(\pm 2\pi i\) or \(4\pi i\) would need poles inside the contour with nonzero residues, and there are none.
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