8.3 Applications

We will cover three applications of Cauchy's integral formula.

Cauchy's Estimate
Liouville's Theorem
Fundamental Theorem of Algebra

Theorem (Cauchy's Estimate)

Fix $f$ holomorphic on $B (b, R)$ . If, for some $0 < r < R$ there is $M > 0$ with $| f (z) | \leq M$ for all $z$ with $| z - b | = r$ then $| f^{(n)} (b) | \leq \frac{M n!}{r^{n}}$ for all $n \in N$ .

Proof:

Our proof of Taylor's theorem gave $f^{(n)} (b) = \frac{n!}{2 π i} \int_{γ (r)} \frac{f (z)}{(z - b)^{n + 1}} d z$ where $γ (t) = b + r e^{i t}$ on $[0, 2 π]$ . The estimation lemma gives $| f^{(n)} (b) | \leq \frac{n!}{2 π} \frac{M}{r^{n + 1}} ℓ (γ (r)) = \frac{M n!}{r^{n}}$ because $| f (b + r e^{i t}) | \leq M$ for all $0 \leq t \leq 2 π$ . $◻$

Definition

A function $f : D \to C$ is bounded if there is $M > 0$ such that $| f (z) | \leq M$ for all $z \in D$ .

Theorem (Liouville's Theorem)

If $f : C \to C$ is holomorphic and bounded then $f$ is constant.

Proof:

Fix $a \in C$ . Since $f$ is holomorphic on all of $C$ we can apply Cauchy's estimate for all $r > 0$ to get $| f^{'} (a) | \leq \frac{M}{r}$ for all $r > 0$ . But this implies $f^{'} (a) = 0$ . This is true for all $a \in C$ so $f$ must be constant. $◻$

Our last application for the moment is the fundamental theorem of algebra. It states that every non-constant polynomial has a root.

Theorem (Fundamental Theorem of Algebra)

Let $p (z) = z^{n} + a_{n - 1} z^{n - 1} + \dots + a_{2} z^{2} + a_{1} z + a_{0}$ be a polynomial. If $p$ is not constant there is $b \in C$ with $p (b) = 0$ .

Proof:

The proof will be by contradiction. Suppose that $p (b) \neq 0$ 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 $lim_{z \to \infty} f (z) = 0$ . But then Liouville's theorem implies $f$ is constant, which contradicts our hypothesis. $◻$