Discussion
The value of the integral
\[ \int_{C} \frac{z^{100}}{z^{101} + 1}\,dz \]
where C is the circle of radius 2 centred at the origin taken in the anti-clockwise direction is
Answer: D
The integral equals \(2\pi i\).
Let \(p(z) = z^{101} + 1\). Then \(p'(z) = 101z^{100}\), so the integrand is \(\frac{1}{101}\frac{p'(z)}{p(z)}\).
All 101 roots of p lie on the unit circle, inside C of radius 2. By the argument principle the contour integral of p'/p counts them:
\[ \begin{aligned} \oint_{C} \frac{z^{100}}{z^{101}+1}\,dz &= \frac{1}{101} \cdot 2\pi i \cdot 101 \\ &= 2\pi i \end{aligned} \]
As a check, on a large circle the integrand behaves like 1/z, whose integral is \(2\pi i\). The value 0 would need no poles inside, \(-2\pi i\) has the wrong orientation, and \(2\pi\) drops the factor i.
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