Cauchy's Integral Formula Derivation
Theorem 1
Let the complex function $f: A \subseteq \mathbb{C} \to \mathbb{C}$ be analytic in a simply connected region $\mathscr{R}$.
If a simple closed path $\mathscr{C} \subset \mathscr{R}$ contained in $\mathscr{R}$ surrounds a point $\alpha$, then the following holds: $$ f(\alpha) = {{1} \over {2 \pi i }} \int_{\mathscr{C}} {{f(z)} \over { z - \alpha }} dz $$
Derivation
First, let’s show that $\displaystyle 2 \pi i = \int_{\mathscr{C} '} {{1} \over { z - \alpha }} dz$.
Contraction auxiliary lemma of complex path integrals: For a circle $\mathscr{C} '$ centered at $\alpha$ inside $\mathscr{C}$, $$\int_{\mathscr{C}} f(z) dz = \int_{\mathscr{C} '} f(z) dz$$
Contracting the integration interval of $\displaystyle \int_{\mathscr{C}} {{1} \over { z - \alpha }} dz$ to the circle $\mathscr{C} ': | z - \alpha | = \rho$ results in $z(\theta) = \rho e^{i \theta} + \alpha, -\pi \le \theta \le \pi$, so $$ \int_{\mathscr{C} '} {{1} \over { z - \alpha }} dz = \int_{-\pi}^{\pi} {{ i \rho e^{i \theta}} \over { \rho e^{i \theta} }} d\theta = 2 \pi i $$ Now, setting $\displaystyle I = \int_{\mathscr{C}} {{f(z)} \over { z - \alpha }} dz$ and calculating $I$ gives $$ \begin{align*} \int_{\mathscr{C}} {{f(z)} \over { z - \alpha }} dz =& \int_{\mathscr{C} '} {{f(\alpha)} \over { z - \alpha }} dz + \int_{\mathscr{C} '} {{f(z) - f(\alpha)} \over { z - \alpha }} dz \\ =& f(\alpha) \int_{\mathscr{C} '} {{1} \over { z - \alpha }} dz + \int_{\mathscr{C} '} {{f(z) - f(\alpha)} \over { z - \alpha }} dz \\ =& f(\alpha) 2 \pi i + \int_{\mathscr{C} '} {{f(z) - f(\alpha)} \over { z - \alpha }} dz \end{align*} $$ Demonstrating that $\displaystyle \int_{\mathscr{C} '} {{f(z) - f(\alpha)} \over { z - \alpha }} dz = 0$ completes the proof.
$f(z)$ being differentiable in $z = \alpha$ means that for some $M>0$, $$ \left| {{f(z) - f(\alpha)} \over { z - \alpha }} \right| \le M $$ Since $\mathscr{C} ' : | z - \alpha | = \rho$, the length of $\mathscr{C} '$ is $2 \pi \rho$.
ML auxiliary lemma: For a positive number $M$ that satisfies $|f(z)| \le M$ and the length of $\mathscr{C}$, $L$, $$ \left| \int_{\mathscr{C}} f(z) dz \right| \le ML $$
According to the ML auxiliary lemma, $$ \left| \int_{\mathscr{C} '} {{f(z) - f(\alpha)} \over { z - \alpha }} dz \right| \le 2 \pi \rho M $$ Now, if we continuously use the contraction auxiliary lemma of complex path integrals around $z = \alpha$, think $$\mathscr{C}_n : | z - \alpha | = \rho_n \\ \mathscr{C}_{n+1} : | z - \alpha | = \rho_{n+1} \\ \rho_{n} > \rho_{n+1} $$ then, when $n \to \infty$, it is $\rho_{n} \to 0$. For all $\rho_{n} >0$, $$ \left| \int_{\mathscr{C}_{n}} {{f(z) - f(\alpha)} \over { z - \alpha }} dz \right| \le 2 \pi \rho_{n} M $$ thus, $$ \left| \int_{\mathscr{C} '} {{f(z) - f(\alpha)} \over { z - \alpha }} dz \right| = 0 $$ Finally, we obtain the following: $$ \int_{\mathscr{C}} {{f(z)} \over { z - \alpha }} dz = f(\alpha) 2 \pi i $$
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Description
It’s the formula that makes the blind see and the lame walk. The mathematical beauty is indescribable, and its usefulness is so profound that its impact is almost beyond measure. Especially regarding integration, it is often called the flower of complex analysis because of the incessant outpouring of rich mathematical results.
Corollary
Meanwhile, the Cauchy integral formula can be generalized for the $n$th derivative. Except for using mathematical induction for the generalization, the proof fundamentally does not differ from that of the Cauchy integral formula. This formula is useful in itself but harbors even more significant implications.
Generalization of Cauchy Integral Formula for Derivatives
Let the function $f: A \subseteq \mathbb{C} \to \mathbb{C}$ be analytic in a simply connected region $\mathscr{R}$.
If a simple closed path $\mathscr{C} \subset \mathscr{R}$ contained in $\mathscr{R}$ surrounds a point $\alpha$, then for a natural number $n$, the following holds: $$ f^{(n)} (\alpha) = {{n!} \over {2 \pi i }} \int_{\mathscr{C}} {{f(z)} \over { (z - \alpha)^{n+1} }} dz $$
However, while reading the conditions, there’s no mentioning that $f$ needs to be differentiable multiple times, yet it uses the $n$th derivative. This means, in complex analysis, a function that is differentiable once is infinitely differentiable. This is guaranteed during the proof process and is a very powerful advantage, which cannot be easily assured in real functions. Thus, complex analysis enables the derivation of incredible mathematical results, as it dismantles various limitations, whether in differentiation or integration.
Infinite Differentiability 2
The derivative of an analytic complex function is analytic. In other words, if $f$ is analytic in $z \in \mathbb{C}$, then for all $n \in \mathbb{N}$, the $n$th order derivative $f^{(n)}$ is also analytic in $z$.