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We will cover three applications of Cauchy's integral formula.
Fix $f$ holomorphic on $\ball(b,R)$. If, for some $0 < r < R$ there is $M > 0$ with $|f(z)| \le M$ for all $z$ with $|z-b| = r$ then \[ |f^{(n)}(b)| \le \dfrac{M n!}{r^n} \] for all $n \in \N$.
Our proof of Taylor's theorem gave \[ f^{(n)}(b) = \dfrac{n!}{2 \pi i} \int\limits_{\gamma(r)} \dfrac{f(z)}{(z-b)^{n+1}} \intd z \] where $\gamma(t) = b + re^{it}$ on $[0,2\pi]$. The estimation lemma gives \[ |f^{(n)}(b)| \le \dfrac{n!}{2\pi} \dfrac{M}{r^{n+1}} \ell(\gamma(r)) = \dfrac{M n!}{r^n} \] because $|f(b+re^{it})| \le M$ for all $0 \le t \le 2\pi$. $\square$
A function $f : D \to \C$ is bounded if there is $M > 0$ such that $|f(z)| \le M$ for all $z \in D$.
If $f : \C \to \C$ is holomorphic and bounded then $f$ is constant.
Fix $a \in \C$. Since $f$ is holomorphic on all of $\C$ we can apply Cauchy's estimate for all $r > 0$ to get \[ |f'(a)| \le \dfrac{M}{r} \] for all $r > 0$. But this implies $f'(a) = 0$. This is true for all $a \in \C$ so $f$ must be constant. $\square$
Our last application for the moment is the fundamental theorem of algebra. It states that every non-constant polynomial has a root.
Let \[p(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_2z^2 + a_1 z + a_0\] be a polynomial. If $p$ is not constant there is $b \in \C$ with $p(b) = 0$.
The proof will be by contradiction. Suppose that $p(b) \ne 0$ for all $b \in \C$. Then the function $f(z) = 1/p(z)$ is holomorphic on $\C$. Moreover, it is bounded because $f$ is continuous and $\lim\limits_{z \to \infty} f(z) = 0$. But then Liouville's theorem implies $f$ is constant, which contradicts our hypothesis. $\square$