We now compute one of the fundamental contour integrals. We will show that $ \int_{\partial B_r(a)}(z-a)^n \mathrm{~d} z= \begin{cases}2 \pi \mathrm{i} & n=-1 \\ 0 & n \neq 1\end{cases} $ where $\partial B_r(a)$ denotes the boundary of the ball of radius $r$, parametrised counter-clockwise (i.e. positively oriented with respect to $\left.B_r(a)\right)$. Observe that the result is uniform with respect to $r$. That is a natural consequence of [[Principle of Deformation of Contours]], given than the functions we are integrating only fail to be analytic at one point (at most, depending on $n$ ). In fact we could have chosen any curve that wraps around $a$ once and obtain the same result. Now, to compute the integral above, notice that we can parametrise the curve as $\gamma(t)=a+r \mathrm{e}^{\mathrm{i} \ell}$, for $t \in[0,2 \pi)$. Therefore we have (since $\gamma^{\prime}(t)=\mathrm{i} r \mathrm{e}^{\mathrm{it}}$ ) $\int_{\partial B_r(a)}(z-a)^n \mathrm{~d} z=\int_0^{2 \pi}\left(r \mathrm{e}^{\mathrm{i} t}\right)^n \mathrm{i} r \mathrm{e}^{\mathrm{i} t} \mathrm{~d} t=\mathrm{i} r^{n+1} \int_0^{2 \pi} \mathrm{e}^{\mathrm{i}(n+1) t} \mathrm{~d} t .$ Notice that in the case $n=-1$ that expression equals $2 \pi \mathrm{i}$. When $n \neq-1$ notice that we obtain 0 , since for all $k \neq 0$ we have $ \int_0^{2 \pi} \mathrm{e}^{\mathrm{i} k t} \mathrm{~d} t=\left.\frac{1}{k} \mathrm{e}^{\mathrm{i} k t}\right|_0 ^{2 \pi}=\frac{1}{k}-\frac{1}{k}=0 $ We restate, in the notation that will be most convenient for the next few results, the fundamental integral above in the case $n=-1$, noting that the result does not depend on $r$. We have $ \int_{\partial B_r(z)} \frac{1}{w-z} \mathrm{~d} w=2 \pi \mathrm{i} $