If $f$ is [[Complex Differentiability|analytic]] in a [[Simple connectedness|simply connected]] domain $D$ and $z_0 \in D$, the quotient $f(w) /\left(w-z\right)$ is undefined at $z$, making it non-analytic in $D$. Consequently, the [[Cauchy-Goursat Theorem|Cauchy-Goursat theorem]] does not allow us to conclude that the integral $\int_C \frac{f(w)}{w-z} d w$around a simple closed contour $C$ containing $w$, is zero. However, as we shall see, the value of this integral is $2 \pi i f\left(z\right)$. This result is the first of two remarkable formulas. Given a simple closed $C^1$ curve $\gamma$ we denote by $I(\gamma)$ the interior region to $\gamma$ so that $\gamma$ is positively oriented (outside is on the right). We denote by $O(\gamma)$ the exterior region to $\gamma$. Notice that by the [[Principle of Deformation of Contours|principle of deformation of contours]] and [[Fundamental Contour Integral]], we have $\int_\gamma \frac{1}{w-z} \mathrm{~d} w=\int_{\partial B_r(z)} \frac{1}{w-z} \mathrm{~d} w=2 \pi \mathrm{i}.\tag{1}$ for every $z \in I(\gamma)$ and every $r$ sufficiently small so that $B_r(z) \subset I(\gamma)$. > [!NOTE] Theorem > Let $\gamma:[a, b] \rightarrow \mathrm{C}$ be a positively oriented simple closed $C^1$ curve. Assume that $f$ is analytic in $\gamma$ and on the interior of $\gamma, I(\gamma)$. Then $\int_\gamma \frac{f(w)}{w-z} \mathrm{~d} w = 2\pi i f(z) \quad \text { for all } z \in I(\gamma).$ **Remark**. This formula has remarkable consequences for analytic functions. First notice that it claims that we can recover the value of $f$ at any point by integration along a curve around that point (provided the curve is sufficiently regular, positively oriented, and contained in $I(\gamma)$ ). This is a very significant difference with respect to smooth functions in $\mathbb{R}^2$ for example. ###### Proof Fix $z \in I(\gamma)$, and choose $r$ small enough so that $B_r(z) \subset I(\gamma)$. By the principle of deformation of contours theorem we have $ \int_\gamma \frac{f(w)}{w-z} \mathrm{~d} w= \int_{\partial B_r(z)} \frac{f(w)}{w-z} \mathrm{~d} w$reducing the problem to considering $\gamma$ as a $\partial B_r(z)$. Observe that the integral is the same for every $r$ sufficiently small, and later on we will exploit this fact by talking limits as $r$ tends to zero. For now, we have $\int_{\partial B_r(z)} \frac{f(w)}{w-z} \mathrm{~d} w=\int_{\partial B_r(z)} \frac{f(z)}{w-z} \mathrm{~d} w+ \int_{\partial B_r(z)} \frac{f(w)-f(z)}{w-z} \mathrm{~d} w=: I+I I .$ Notice that the first integral $I$ equals $f(z)$. Indeed, using $(1)$ $ \int_{\partial B_r(z)} \frac{f(z)}{w-z} \mathrm{~d} w=f(z) \int_{\partial B_r(z)} \frac{1}{w-z} \mathrm{~d} w= 2\pi \mathrm{i} f(z).$ All that remains to is to show that $I I=0$. Notice that since $f$ is analytic in $I(\gamma)$, given any $\varepsilon>0$ we can find $r$ sufficiently small so that $|f(w)-f(z)| \leqslant \varepsilon \quad \text { for all } w \in \partial B_r(z)$ We parametrise $\partial B_r(z)$ counterclockwise by $\gamma(t)=z+r \mathrm{e}^{\mathrm{i} t}$ for $t \in[0,2 \pi)$. We have $\gamma^{\prime}(t)=\mathrm{ir} \mathrm{e}^{\mathrm{i} t}$ and therefore $\begin{align} |I I|&=\left| \int_{\partial B_r(z)} \frac{f(w)-f(z)}{w-z} \mathrm{~d} w\right| \\ &= \left|\int_{0}^{2\pi} \frac{f(z+re^{it})-f(z)}{re^{it}} rie^{it} \, dw \right | \\ & \int_0^{2 \pi}\left|f\left(z+r \mathrm{e}^{\mathrm{i} t}\right)-f(z)\right| \mathrm{d} t \leqslant \varepsilon \end{align}$ Since $\varepsilon$ is arbitrary we obtain the desired result.