Analytic ⟹ Infinitely Differentiable

Cauchy's integral formula expresses a holomorphic function as an integral whose only dependence on a sits in the tidy factor 1/(z - a). That factor can be differentiated as often as we please — and with it comes the most startling theorem of complex calculus: differentiable once means differentiable infinitely often.

Differentiating under the integral sign

Start from the integral formula and differentiate with respect to a. The contour and f(z) do not depend on a, so only 1/(z - a) is touched.

Step 1 — differentiate the kernel once. Since \dfrac{\partial}{\partial a}\dfrac{1}{z - a} = \dfrac{1}{(z - a)^2}:

f'(a) = \frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{(z - a)^2}\,dz.

Step 2 — differentiate again. Each derivative drops the power by one and multiplies by the old power:

f''(a) = \frac{2!}{2\pi i}\oint_\gamma \frac{f(z)}{(z - a)^3}\,dz.

Step 3 — the general pattern. Differentiating n times brings out a factor of n! and raises the denominator to n + 1:

f^{(n)}(a) = \frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z - a)^{\,n+1}}\,dz.

The integral on the right exists for every n, so every derivative exists. A function that is complex-differentiable once is automatically smooth to all orders.

If f is holomorphic, then it has derivatives of all orders, given by \displaystyle f^{(n)}(a) = \frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z - a)^{\,n+1}}\,dz.
One complex derivative forces infinitely many — utterly unlike real calculus, where a once-differentiable function may have no second derivative at all.

Liouville's theorem: bounded entire ⟹ constant

A function holomorphic on the whole plane is called entire. The n = 1 formula gives a clean bound on its derivative, and that bound proves a striking rigidity.

Step 1 — bound f' on a big circle. Take \gamma to be the circle of radius R about a, and suppose |f| \le M everywhere. On that circle |z - a| = R, so the integrand of f' has size at most M/R^2. The ML inequality with L = 2\pi R gives

|f'(a)| \le \frac{1}{2\pi}\cdot\frac{M}{R^2}\cdot 2\pi R = \frac{M}{R}.

Step 2 — let the radius grow. Because f is entire, this holds for every R. Send R \to \infty:

|f'(a)| \le \frac{M}{R} \;\longrightarrow\; 0.

Step 3 — conclude. So f'(a) = 0 at every point a. A function with zero derivative everywhere is constant.

A bounded entire function is constant.
The proof is one ML estimate, |f'(a)| \le M/R, pushed to the limit R \to \infty.

Liouville's theorem cracks a classic problem in one stroke. Suppose a non-constant polynomial p(z) had no root. Then 1/p(z) would be holomorphic on the whole plane — entire. As |z| \to \infty a polynomial grows without bound, so 1/p(z) \to 0; combined with continuity, that makes 1/p bounded. A bounded entire function is constant by Liouville, so 1/p — and hence p — would be constant, contradicting the assumption. Therefore every non-constant polynomial has a root: the fundamental theorem of algebra, delivered by complex integration.

Watch the derivative bound vanish

Fix a bound M on a bounded entire function and grow the radius R. The circle is the contour in the n = 1 formula; the readout tracks the Liouville bound |f'(a)| \le M/R. As R grows the ceiling falls toward zero — and a derivative bounded by something heading to zero must itself be zero.