Solution: Let \[ f(z) = z^{101} + 1\implies f'(z) = 101 z^{100} . \] We recall the argument principal which says that if $f(z)$ is analytic on and inside a simple closed curve $C$ (oriented counterclockwise) except for (possibly) some finite poles inside (not on) $C$ and some zeros inside (not on) $C$. Then, \[ \frac{1}{2\pi \iota } \oint_{C} \frac{f'(z)}{f(z)} \mathrm{d} z = Z - P, \] where $Z$ is the number of poles (with multiplicity) of $f$ inside $C$ and $Z$ is the zeros (with multiplicity) of $f$ inside $C$.

In this problem, we have \begin{align*} \frac{1}{2\pi \iota } \oint_{\vert z \vert = 2} \frac{f'(z)}{f(z)} \mathrm{d} z & = \frac{1}{2\pi \iota } \oint_{\vert z \vert = 2} \frac{101 z^{100}}{z^{101} + 1} \mathrm{d} z = Z - P. \end{align*} Since $f$ is a polynomial of degree $101$, it will have $101$ zeros and no poles. Also, all the zeros will lie inside the given circle $\vert z \vert =2$ as all the roots are of unit length. Thus, \begin{align*} \frac{1}{2\pi \iota } \oint_{\vert z \vert = 2} \frac{101 z^{100}}{z^{101} + 1} \mathrm{d} z = 101 - 0 & \implies \oint_{C} \frac{z^{100}}{z^{101} + 1} \mathrm{d} z = 2\pi \iota . \end{align*}