Recall the Cauchy integral formula: If f is analytic on and inside the simple closed curve C, traversed positively, and z_0 \in \operatorname{Int} C, then f(z_0) = \frac 1 {2\pi i}\int_C \frac{f(z)}{z-z_0}\,dz. Theorem. Under the same conditions, f^{(n)}(z_0) = \frac{n!}{2\pi i}\int_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz. Proof. See exercise 9 of §57 (8 Ed §52). \square
As a result, this tell us that for f = u+iv and f analytic at z_0 = x_0 + iy_0, we know that partials of all orders of u and v exist and are continuous at (x_0, y_0). This is very different from the situation in \mathbb R, where it is very easy to have functions with continuous derivatives but not differentiable. For example, with f(x) = |x|^3, f, f' and f'' are continuous but f'''(0) does not exist.
Note: If f is analytic at z_0, then its derivatives of all orders exist and are analytic at z_0.
Theorem (Morera). Let f be continuous on a domain \Omega. If \int_C f(z)\,dz = 0 for all closed contours C in \Omega, then f is analytic on \Omega.
Proof. By the theorem from lecture 19, f has a primitive F because \int_Cf(Z)\,dz = 0. But then, F' = f exists and is continuous on \Omega by assumption of the theorem. This tell us that F is analytic. Hence, by the note above, f = F' is also analytic. \square
A number of nice results follow from the theorem with f^{(n)}(z_0) above.
Result (I). Let f be analytic in and on C_R(z_0) (curve of a circle of radius R around z_0) and set M_R = \max_{z \in C_R}|f(z)|. Then, \left|f^{(n)}(z_0)\right| \le \frac{n!M_R}{R^n}. This tells us that if we know what the function does on the circle, we can estimate the size of its derivatives at a point. In fact, the closer we get, the worse this estimate becomes because of the division by R^n.
Proof. M_R is well defined by the extreme value theorem. Then, applying the aforementioned theorem, \begin{aligned} \left|f^{(n)}(z_n)\right| = \left|\frac{n!}{2\pi i} \int_{C_R}\frac{f(z)}{(z-z_0)^{n+1}}\,dz\right| &\le \frac{n!}{2\pi}\int_{C_R}\frac{|f(z)|}{|z-z_0|^{n+1}}\,dz \\ &\le \frac{n!M_R}{2\pi R^{n+1}}\int_{C_R}dz \\ &= \frac{n!M_R}{R^n} \end{aligned} Above, note that |z-z_0|=R on this contour, and \int_{C_R}dz is just the arc length of C_R (equal to 2\pi R). \square
As a brief discussion, we have all these powerful results about analytic functions in \mathbb C. However, this hints that being complex differentiable is actually a very restrictive condition.
Result (II – Liouville). If f : \mathbb C \to \mathbb C is bounded and entire (everywhere differentiable), then f is constant.
Proof. Suppose |f| \le M on all of \mathbb C and it is entire. Apply result I for n=1 on C_R(z_0), an arbitrary circle around z_0. The result implies that |f'(z_0)| \le \frac{1!M}R = \frac M R. Letting R \to \infty, we see that f'(z_0) = 0. Since z_0 was arbitrary, we have the result. \square
This is clearly not the case in \mathbb R.
Result (III – Fundamental theorem of algebra). An n-th degree polynomial has exactly n zeros.