Question

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

• A. \(2\pi i\)
• B. \(-2\pi i\)
• C. \(0\)
• D. \(4\pi i\)

Hint:

Factor \(1-z^{101}\) as \((1-z)(1+z+\cdots+z^{100})\) and check whether any zero of the simplified denominator lies inside \(|z|=\frac{7}{9}\).

Answer 1

The Correct Option is C

A different way to see this is to expand the integrand directly as a convergent power series inside the disk bounded by $\Gamma$, instead of factoring out roots of unity.

Since $\Gamma$ has radius $\frac{7}{9}<1$, every point $z$ inside or on $\Gamma$ satisfies $|z|\le\frac{7}{9}<1$. For such $z$, the geometric series

\[ \frac{1}{1-z^{101}} = \sum_{k=0}^{\infty} z^{101k} \]

converges, because $|z^{101}|=|z|^{101}<1$.

The function $3\cos z + 5\sin z$ is entire, so it also has a convergent Taylor series for every $z$, in particular for $|z|\le \frac{7}{9}$.

Multiplying $(1-z)$, the entire function $3\cos z+5\sin z$, and the convergent geometric series $\sum z^{101k}$ together gives a power series in $z$ that converges for all $|z|<1$. A function represented by a convergent power series on a disk is analytic throughout that disk.

So the whole integrand is analytic on the closed disk $|z|\le \frac{7}{9}$ enclosed by $\Gamma$, with no singularities inside at all.

Let's summarize:

• Writing $\frac{1}{1-z^{101}}$ as a geometric series shows the integrand is just an ordinary convergent power series inside $|z|<1$.
• Cauchy's theorem says the integral of an analytic function around any closed contour lying in its domain of analyticity is zero.

Since $\Gamma$ lies entirely inside the region of analyticity, the integral must be 0, option (C).

\[ \boxed{0} \]