Derivation of Cauchy's Integral Formula
Theorem 1
Let a complex function $f: A \subseteq \mathbb{C} \to \mathbb{C}$ be analytic on a simply connected region $\mathscr{R}$.
If a simple closed path $\mathscr{C} \subset \mathscr{R}$ contained in $\mathscr{R}$ encloses some 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 us show that $\displaystyle 2 \pi i = \int_{\mathscr{C} '} {{1} \over { z - \alpha }} dz$.
Deformation lemma for complex contour 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 path of $\displaystyle \int_{\mathscr{C}} {{1} \over { z - \alpha }} dz$ to the circle $\mathscr{C} ': | z - \alpha | = \rho$, since $z(\theta) = \rho e^{i \theta} + \alpha, -\pi \le \theta \le \pi$, $$ \int_{\mathscr{C} '} {{1} \over { z - \alpha }} dz = \int_{-\pi}^{\pi} {{ i \rho e^{i \theta}} \over { \rho e^{i \theta} }} d\theta = 2 \pi i $$ Now, setting $\displaystyle I = \int_{\mathscr{C}} {{f(z)} \over { z - \alpha }} dz$ and computing $I$, $$ \begin{align*} \int_{\mathscr{C}} {{f(z)} \over { z - \alpha }} dz =& \int_{\mathscr{C} '} {{f(\alpha)} \over { z - \alpha }} dz + \int_{\mathscr{C} '} {{f(z) - f(\alpha)} \over { z - \alpha }} dz \\ =& f(\alpha) \int_{\mathscr{C} '} {{1} \over { z - \alpha }} dz + \int_{\mathscr{C} '} {{f(z) - f(\alpha)} \over { z - \alpha }} dz \\ =& f(\alpha) 2 \pi i + \int_{\mathscr{C} '} {{f(z) - f(\alpha)} \over { z - \alpha }} dz \end{align*} $$ The proof is complete once we show that $\displaystyle \int_{\mathscr{C} '} {{f(z) - f(\alpha)} \over { z - \alpha }} dz = 0$.
Since $f(z)$ is differentiable at $z = \alpha$, 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 lemma: For a positive number $M$ satisfying $|f(z)| \le M$ and the length $L$ of $\mathscr{C}$, $$ \left| \int_{\mathscr{C}} f(z) dz \right| \le ML $$
By the ML lemma, $$ \left| \int_{\mathscr{C} '} {{f(z) - f(\alpha)} \over { z - \alpha }} dz \right| \le 2 \pi \rho M $$ Now let us think of continuously applying the deformation lemma for complex contour integrals around $z = \alpha$. That is, $$\mathscr{C}_n : | z - \alpha | = \rho_n \\ \mathscr{C}_{n+1} : | z - \alpha | = \rho_{n+1} \\ \rho_{n} > \rho_{n+1} $$ if this holds, then as $n \to \infty$, $\rho_{n} \to 0$. Since for all $\rho_{n} >0$ $$ \left| \int_{\mathscr{C}_{n}} {{f(z) - f(\alpha)} \over { z - \alpha }} dz \right| \le 2 \pi \rho_{n} M $$ holds, $$ \left| \int_{\mathscr{C} '} {{f(z) - f(\alpha)} \over { z - \alpha }} dz \right| = 0 $$ Finally, we obtain the following. $$ \int_{\mathscr{C}} {{f(z)} \over { z - \alpha }} dz = f(\alpha) 2 \pi i $$
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Explanation
This is a formula that makes the blind see and the lame leap up. Not to mention its mathematical beauty, it is so useful that the shock it delivers is almost beyond measure. In particular, regarding integration, rich mathematical results pour out endlessly, so it is even called the flower of complex analysis.
Corollary
Meanwhile, Cauchy’s integral formula can be generalized to the $n$th derivative. The proof is essentially no different from that of Cauchy’s integral formula, except that it uses mathematical induction for the generalization. This formula is extremely useful in its own right, but it also carries an even more important meaning.
Cauchy’s Integral Formula Generalized to Derivatives
Let a function $f: A \subseteq \mathbb{C} \to \mathbb{C}$ be analytic on a simply connected region $\mathscr{R}$.
If a simple closed path $\mathscr{C} \subset \mathscr{R}$ contained in $\mathscr{R}$ encloses some 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, reading the condition, there is no statement that $f$ must be differentiable several times, yet the $n$th derivative is being used. That is, in complex analysis it means that a function that is differentiable once is differentiable infinitely many times. This is guaranteed in the course of the proof and is a very, very powerful advantage, a property that cannot be easily guaranteed for real functions. In this way, complex analysis breaks down all sorts of restrictions, whether in differentiation or integration, so that astonishing mathematical results can be deduced with ease.
Infinite Differentiability 2
The derivative of a complex function is analytic. In other words, if $f$ is analytic at $z \in \mathbb{C}$, then for all $n \in \mathbb{N}$ the $n$th derivative $f^{(n)}$ is also analytic at $z$.