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

Step 1: Simplify the integrand using a factorization.
Recall the identity \(1-z^{n} = (1-z)(1+z+z^2+\cdots+z^{n-1})\). With \(n=101\), this gives \(1-z^{101} = (1-z)(1+z+z^2+\cdots+z^{100})\).
So the integrand becomes
\[ \frac{(1-z)(3\cos z+5\sin z)}{(1-z)(1+z+\cdots+z^{100})} = \frac{3\cos z+5\sin z}{1+z+z^2+\cdots+z^{100}} \] after cancelling the common factor \((1-z)\); this cancellation turns the apparent singularity at \(z=1\) into a removable one.

Step 2: Locate the singularities of the simplified integrand.
The numerator \(3\cos z+5\sin z\) is entire. The denominator \(1+z+z^2+\cdots+z^{100}\) is zero exactly at the 101th roots of unity other than \(z=1\), because \((1-z)\) times it equals \(1-z^{101}\). Every such root satisfies \(|z|=1\).

Step 3: Compare with the contour.
The contour \(\Gamma\) is the circle \(|z|=\frac{7}{9}\), and \(\frac{7}{9}<1\). Since all 100 zeros of the denominator lie exactly on \(|z|=1\), none of them lie inside or on \(\Gamma\). The removable point \(z=1\) is also outside \(\Gamma\), since \(|1|=1>\frac{7}{9}\).
So the simplified integrand is analytic on and inside \(\Gamma\), with no singular points enclosed.

Step 4: Apply Cauchy's theorem and check the wrong options.
By the Cauchy-Goursat theorem, the integral of a function that is analytic everywhere inside and on a closed contour is zero. Options (A), (B) and (D) would all require some singularity to be enclosed by \(\Gamma\) so that a residue contributes a nonzero multiple of \(\pi i\), but there is no such singularity, so they are wrong.

Final Answer:
The contour integral equals 0. \[ \boxed{0} \]