> [!NOTE] Theorem > Let $f$ be an [[Complex Differentiability|analytic function]] on $B_{R}(a)$ for $a\in \mathbb{C}$, $R>0$. Then there exists unique constants $c_{n}$, $n\in \mathbb{N}$, such that $f(z)=\sum_{n=0}^\infty c_{n} (z-a)^{n}, \quad \forall z\in B_{R}(a).$Moreover, the coefficients $c_{n}$ are given by $c_{n} = \frac{1}{2\pi \mathrm{i}}\oint_{\gamma} \frac{f(w)}{(w-a)^{n+1}} \, dw = \frac{f^{(n)}(a)}{n!},$where $\gamma$ is a positively oriented simple closed curve (piece-wise $C^{1}$) that is contained in $B_{R}(a)$ with $a\in I(\gamma)$, the interior of $\gamma$. ###### Proof Given some $z \in B_R(a)$ we will take $\gamma$ to be $\partial B_r(a)$ (positively oriented), for $r$ small enough so that $|z-a|<r<R$. We can use the Theorem of deformation of contours to prove the integrals over all curves $\gamma$ as above are the same. By [[Cauchy's Integral Formula]] $f(z)=\frac{1}{2 \pi \mathrm{i}} \int_{\partial B_r(a)} \frac{f(w)}{w-z} \mathrm{~d} w \tag{1}$ Notice that since $|w-a|=r$ and we have chosen $r$ so that $|z-a|<r$ we have $|z-a|<|w-a|$ for all $w \in \partial B_r(a)$. As a result $ \frac{|z-a|}{|w-a|}<1 $ and we can use the geometric series expansion to obtain $ \frac{1}{w-z}=\frac{1}{w-a} \frac{1}{\left(1-\frac{z-a}{w-a}\right)}=\frac{1}{w-a} \sum_{n=0}^{\infty}\left(\frac{z-a}{w-a}\right)^n $ Inserting this expression in (1) we obtain $ f(z)=\frac{1}{2 \pi \mathrm{i}} \int_{\partial B_r(a)} f(w) \frac{1}{w-a} \sum_{n=0}^{\infty}\left(\frac{z-a}{w-a}\right)^n \mathrm{~d} w $ For $w \in \partial B_r(a)$ the series converges absolutely (Weierstrass M -test), and therefore we can exchange the order of the summation and integration to obtain $ f(z)=\sum_{n=0}^{\infty} \frac{1}{2 \pi \mathrm{i}} \int_{\partial B_r(a)} \frac{f(w)}{(w-a)^{n+1}} \mathrm{~d} w(z-a)^n=\sum_{n=0}^{\infty} c_n(z-a)^n $ obtaining the desired result. It remains to show that the coefficients are unique. Now, assume that $f(z)=\sum_{k=0}^{\infty} b_k(z-a)^k$ for some $b_k \in \mathbb{C}$. We have $ \begin{gathered} \int_{\partial B_r(a)} \frac{f(w)}{(w-a)^{n+1}} \mathrm{~d} w=\int_{\partial B_r(a)} \sum_{k=0}^{\infty} b_k(w-a)^k \frac{1}{(w-a)^{n+1}} \mathrm{~d} w \\ =\sum_{k=0}^{\infty} b_k \int_{\partial B_r(a)}(w-a)^{k-n-1} \mathrm{~d} w=2 \pi \mathrm{i} b_n \end{gathered} $ where we have used the fundamental integrals, together with the fact that we can commute the summation and integration. This proves that $b_n=c_n$, concluding the proof.