Cauchy's formula for derivatives, combined with the ML inequality, limits how fast an analytic function can change. For a function analytic on the whole plane, the limit becomes so strong that a bounded one cannot change at all. One consequence finishes a promise from Algebra 2: every polynomial has a complex root.
Key idea
Cauchy's estimate. If $f$ is analytic on and inside $|z - a| = R$ and $|f| \le M$ on that circle, then $|f^{(n)}(a)| \le \dfrac{n!\,M}{R^n}$.
Liouville's theorem. A bounded entire function is constant.
Fundamental theorem of algebra. Every nonconstant polynomial has a complex zero, so a polynomial of degree $n$ has exactly $n$ zeros counted with multiplicity.
Liouville's theorem. A bounded entire function is constant.
Fundamental theorem of algebra. Every nonconstant polynomial has a complex zero, so a polynomial of degree $n$ has exactly $n$ zeros counted with multiplicity.
Why it works
By the derivative formula and ML: $|f^{(n)}(a)| = \frac{n!}{2\pi}\left|\oint \frac{f(z)}{(z - a)^{n + 1}}dz\right| \le \frac{n!}{2\pi}\cdot\frac{M}{R^{n + 1}}\cdot 2\pi R = \frac{n!M}{R^n}$. If $f$ is entire with $|f| \le M$ everywhere, take $n = 1$ and let $R \to \infty$: $|f'(a)| \le \frac{M}{R} \to 0$, so $f' = 0$ everywhere and $f$ is constant. For a polynomial $p$ with no zero, $\frac{1}{p}$ would be entire and tend to $0$ as $|z| \to \infty$, hence bounded, hence constant, which is impossible unless $p$ is constant. Dividing out one root at a time gives all $n$.
Worked examples
Worked example: A bounded entire function
An entire function satisfies $|f(z)| \le 7$ for all $z$ and $f(2) = 3 - i$. Find $f(5i)$.
- $f$ is entire and bounded, so it is constant.
- $f(5i) = f(2) = 3 - i$.
Answer: $3 - i$
Worked example: Cauchy's estimate
$f$ is analytic on and inside $|z| = 3$ with $|f| \le 9$ there. How large can $|f''(0)|$ be?
- $|f''(0)| \le \frac{2!\cdot 9}{3^2} = 2$.
- The bound is reached by $f(z) = z^2$, which has $|f| = 9$ on the circle and $f''(0) = 2$.
Answer: at most $2$
Worked example: Growth forces a polynomial
An entire function has $|f(z)| \le 3|z|^2 + 5$ for all $z$. What can you say about $f'''(0)$?
- Cauchy's estimate on $|z| = R$: $|f'''(0)| \le \frac{3!\,(3R^2 + 5)}{R^3}$.
- As $R \to \infty$ the right side tends to $0$, so $f'''(0) = 0$; the same works for every higher derivative at every point.
- So $f$ is a polynomial of degree at most $2$.
Answer: $f'''(0) = 0$
Common mistake
Offering $\sin z$ (or $\cos z$, or $e^{iz}$) as a bounded entire function. They are bounded on the real line, but not on the plane: $|\sin(iy)| = |\sinh y|$ and $|e^{iz}| = e^{-y}$ both blow up. Liouville's theorem has no exceptions. When a function seems to break it, check whether it is really entire (no poles anywhere) and really bounded on all of $\mathbb{C}$, especially along the imaginary axis.