0.1: Motivation, Single Variable, and Cauchy's Formula

We start with some standard notation. We use $\mathbb{C}$ for complex numbers, $\mathbb{R}$ for real numbers, $\mathbb{Z}$ for integers, $\mathbb{N} = \{ 1,2,3,\ldots \}$ for natural numbers, $i = \sqrt{-1}$. Throughout this book, the standard terminology of domain means a connected open set. We try to avoid using it if connectedness is not needed, but sometimes we use it just for simplicity.

As complex analysis deals with complex numbers, perhaps we should begin with $\sqrt{-1}$. Start with the real numbers, $\mathbb{R}$, and add $\sqrt{-1}$ into our field. Call this square root $i$, and write the complex numbers, $\mathbb{C}$, by identifying $\mathbb{C}$ with $\mathbb{R}^2$ using $$z = x+iy,$$ where $z \in \mathbb{C}$, and $(x,y) \in \mathbb{R}^2$. A subtle philosophical issue is that there are two square roots of $-1$. Two chickens are running around in our yard, and because we like to know which is which, we catch one and write “$i$” on it. If we happened to have caught the other chicken, we would have gotten an exactly equivalent theory, which we could not tell apart from the original.

Given a complex number $z$, its “opposite” is the complex conjugate of $z$ and is defined as $$\bar{z} \overset{\text{def}}{=} x-iy.$$ The size of $z$ is measured by the so-called modulus, which is just the Euclidean distance: $$|{z}| \overset{\text{def}}{=} \sqrt{z \overline{z}} = \sqrt{x^2+y^2} .$$

If $z = x+iy \in \mathbb{C}$ for $x,y \in \mathbb{R}$, then $x$ is called the real part and $y$ is called the imaginary part. We write $$\Re z = \Re (x+iy) = \frac{z+\bar{z}}{2} = x, \qquad \Im z = \Im (x+iy) = \frac{z-\bar{z}}{2i} = y .$$

A function $f \colon U \subset \mathbb{R}^n \to \mathbb{C}$ for an open set $U$ is said to be continuously differentiable, or $C^1$ if the first (real) partial derivatives exist and are continuous. Similarly, it is $C^k$ or $C^k$-smooth if the first $k$ partial derivatives all exist and are differentiable. Finally, a function is said to be $C^\infty$ or simply smooth$^{1}$ if it is infinitely differentable, or in other words, if it is $C^k$ for all $k \in \mathbb{N}$.

Complex analysis is the study of holomorphic (or complex-analytic) functions. Holomorphic functions are a generalization of polynomials, and to get there one leaves the land of algebra to arrive in the realm of analysis. One can do an awful lot with polynomials, but sometimes they are just not enough. For example, there is no nonzero polynomial function that solves the simplest of differential equations, $f' = f$. We need the exponential function, which is holomorphic.

We start with polynomials. A polynomial in $z$ is an expression of the form $$P(z) = \sum_{k=0}^d c_k \, z^k ,$$ where $c_k \in \mathbb{C}$ and $c_d \not= 0$. The number $d$ is called the degree of the polynomial $P$. We can plug in some number $z$ and compute $P(z)$, to obtain a function $P \colon \mathbb{C} \to \mathbb{C}$.

We try to write $$f(z) = \sum_{k=0}^\infty c_k \, z^k$$ and all is very fine until we wish to know what $f(z)$ is for some number $z \in \mathbb{C}$. We usually mean $$\sum_{k=0}^\infty c_k \, z^k = \lim_{d\to\infty} \sum_{k=0}^d c_k \, z^k .$$ As long as the limit exists, we have a function. You know all this; it is your one-variable complex analysis. We usually start with the functions and prove that we can expand into series.

Let $U \subset \mathbb{C}$ be open. A function $f \colon U \to \mathbb{C}$ is holomorphic (or complex-analytic) if it is complex-differentiable at every point, that is, if $$f'(z) = \lim_{\xi \in \mathbb{C} \to 0} \frac{f(z+\xi)-f(z)}{\xi} \qquad \text{exists for all z } \in\text{ U.}$$ Importantly, the limit is taken with respect to complex $\xi$. Another vantage point is to start with a continuously differentiable$^{2}$ $f$, and say $f = u + i\, v$ is holomorphic if it satisfies the Cauchy-Riemann equations: $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} , \qquad \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} .$$ The so-called Wirtinger operators, $$\frac{\partial}{\partial z} \overset{\text{def}}{=} \frac{1}{2} \left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right), ~ ~ ~ ~ ~ \frac{\partial}{\partial \bar{z}} \overset{\text{def}}{=} \frac{1}{2} \left( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right) ,$$ provide an easier way to understand the Cauchy–Riemann equations. These operators are determined by insisting $$\frac{\partial}{\partial z} z = 1, \quad \frac{\partial}{\partial z} \bar{z} = 0, \quad \frac{\partial}{\partial \bar{z}} z = 0, \quad \frac{\partial}{\partial \bar{z}} \bar{z} = 1.$$

The function $f$ is holomorphic if and only if $$\frac{\partial f}{\partial \bar{z}} = 0 .$$ That seems a far nicer statement of the Cauchy–Riemann equations; it is just one complex equation. It says a function is holomorphic if and only if it depends on $z$ but not on $\bar{z}$ (perhaps that does not make a whole lot of sense at first glance). We check: $$\frac{\partial f}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} \right) = \frac{1}{2} \left( \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} + i \frac{\partial u}{\partial y} - \frac{\partial v}{\partial y} \right) = \frac{1}{2} \left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} \right) + \frac{i}{2} \left( \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} \right) .$$ This expression is zero if and only if the real parts and the imaginary parts are zero. In other words, $$\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} = 0, \qquad \text{and} \qquad \frac{\partial v}{\partial x} + \frac{\partial u}{\partial y} = 0 .$$ That is, the Cauchy–Riemann equations are satisfied.

If $f$ is holomorphic, the derivative in $z$ is the standard complex derivative you know and love: $$\frac{\partial f}{\partial z} (z_0) = f'(z_0) = \lim_{\xi \to 0} \frac{f(z_0+\xi)-f(z_0)}{\xi} .$$ That is because $$\begin{align}\begin{aligned} \frac{\partial f}{\partial z} = \frac{1}{2} \left( \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} \right) + \frac{i}{2} \left( \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} \right) & = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{\partial f}{\partial x} \\ & = \frac{1}{i} \left( \frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y} \right) = \frac{\partial f}{\partial (iy)} . \end{aligned}\end{align}$$

A function on $\mathbb{C}$ is a function defined on $\mathbb{R}^2$ as identified above and so it is a function of $x$ and $y$. Writing $x = \frac{z+\bar{z}}{2}$ and $y = \frac{z-\bar{z}}{2i}$, think of it as a function of two complex variables, $z$ and $\bar{z}$. Pretend for a moment as if $\bar{z}$ did not depend on $z$. The Wirtinger operators work as if $z$ and $\bar{z}$ really were independent variables. For instance: $$\frac{\partial}{\partial z} \left[ z^2 \bar{z}^3 + z^{10} \right] = 2z \bar{z}^3 + 10 z^{9} \qquad \text{and} \qquad \frac{\partial}{\partial \bar{z}} \left[ z^2 \bar{z}^3 + z^{10} \right] = z^2 ( 3 \bar{z}^2 ) + 0 .$$ A holomorphic function is a function “not depending on $\bar{z}$.”

The most important theorem in one variable is the Cauchy integral formula.

The Cauchy formula is the essential ingredient we need from one complex variable. It follows from Green’s theorem$^{3}$ (Stokes’ theorem in two dimensions). You can look forward to Theorem 4.1.1 for a proof of a more general formula, the Cauchy–Pompeiu integral formula.

As a differential form, $dz = dx + i \, dy$. If you are uneasy about differential forms, you probably defined the path integral above directly using the Riemann–Stieltjes integral in your one-complex-variable class. Let us write down the formula in terms of the standard Riemann integral in a special case. Take the unit disc $$\mathbb{D} \overset{\text{def}}{=} \bigl\{ z \in \mathbb{C} : |z| < 1 \bigr\} .$$ The boundary is the unit circle $\partial \mathbb{D} = \bigl\{ z \in \mathbb{C} : |z| = 1 \bigr\}$ oriented positively, that is, counter-clockwise. Parametrize $\partial \mathbb{D}$ by $e^{it}$, where $t$ goes from 0 to $2\pi$. If $\zeta = e^{it}$, then $d\zeta = ie^{it}dt$, and $$f(z) = \frac{1}{2\pi i} \int_{\partial \mathbb{D}} \frac{f(\zeta)}{\zeta-z} \, d \zeta = \frac{1}{2\pi} \int_0^{2\pi} \frac{f(e^{it}) e^{it} }{e^{it}-z} \, dt .$$ If you are not completely comfortable with path integrals, try to think about how you would parametrize the path, and write the integral as an integral any calculus student would recognize.

I venture a guess that 90% of what you learned in a one-variable complex analysis course (depending on who taught it) is more or less a straightforward consequence of the Cauchy integral formula. An important theorem from one variable that follows from the Cauchy formula is the maximum modulus principle (or just the maximum principle). Let us give its simplest version.

That is, if the supremum is attained in the interior of the domain, then the function must be constant. Another way to state the maximum principle is to say: If $f$ extends continuously to the boundary of a domain, then the supremum of $|f(z)|$ is attained on the boundary. In one variable you learned that the maximum principle is really a property of harmonic functions.

In one variable, if $f = u+iv$ is holomorphic for real valued $u$ and $v$, then $u$ and $v$ are harmonic. Locally, any harmonic function is the real (or imaginary) part of a holomorphic function, so in one complex variable, studying harmonic functions is almost equivalent to studying holomorphic functions. Things are decidedly different in two or more variables.

Holomorphic functions admit a power series representation in $z$ at each point $a$: $$f(z) = \sum_{k=0}^\infty c_k {(z-a)}^k .$$ No $\bar{z}$ is necessary, since $\frac{\partial f}{\partial \bar{z}} = 0$.

Let us see the proof using the Cauchy integral formula as we will require this computation in several variables as well. Given $a \in \mathbb{C}$ and $\rho > 0$, define the disc of radius $\rho$ around $a$ $$\Delta_\rho(a) \overset{\text{def}}{=} \bigl\{ z \in \mathbb{C} : |z-a| < \rho \bigr\} .$$ Suppose $U \subset \mathbb{C}$ is open, $f \colon U \to \mathbb{C}$ is holomorphic, $a \in U$, and $\overline{\Delta_\rho(a)} \subset U$ (that is, the closure of the disc is in $U$, and so its boundary $\partial \Delta_\rho(a)$ is also in $U$).

For $z \in \Delta_\rho(a)$ and $\zeta \in \partial \Delta_\rho(a)$, $$\left|\frac{z-a}{\zeta-a}\right| = \frac{|z-a|}{\rho} < 1 .$$ In fact, if $|z-a| \leq \rho' < \rho$, then $\left|\frac{z-a}{\zeta-a}\right| \leq \frac{\rho'}{\rho} < 1$. Therefore, the geometric series $$\sum_{k=0}^\infty {\left(\frac{z-a}{\zeta-a}\right)}^k = \frac{1}{1- \frac{z-a}{\zeta-a}} = \frac{\zeta-a}{\zeta-z}$$ converges uniformly absolutely for $(z,\zeta) \in \overline{\Delta_{\rho'}(a)} \times \partial \Delta_\rho(a)$ (that is, $\sum_k {\bigl\lvert \frac{z-a}{\zeta-a} \bigr\rvert}^k$ converges uniformly).

Let $\gamma$ be the path going around $\partial \Delta_\rho(a)$ once in the positive direction. Compute

$$\begin{align}\begin{aligned} f(z)&=\frac{1}{2\pi i}\int_{\gamma} \frac{f(\zeta )}{\zeta -z}d\zeta \\ &=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta )}{\zeta -a}\frac{\zeta -a}{\zeta -z}d\zeta \\ &=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta )}{\zeta -a}\sum\limits_{k=0}^\infty \left(\frac{z-a}{\zeta -a}\right)^{k}d\zeta \\ &=\sum\limits_{k=0}^\infty \left(\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta )}{(\zeta -a)^{k+1}}d\zeta\right) (z-a)^{k}.\end{aligned}\end{align}$$ In the last equality, we may interchange the limit on the sum with the integral either via Fubini’s theorem or via uniform convergence: $z$ is fixed and if $M$ is the supremum of $\left|\frac{f(\zeta)}{\zeta-a}\right| = \frac{|f(\zeta)|}{\rho}$ on $\partial \Delta_\rho(a)$, then $$\left| \frac{f(\zeta)}{\zeta-a} {\left(\frac{z-a}{\zeta-a}\right)}^k \right| \leq M {\left(\frac{|z-a|}{\rho}\right)}^k, \qquad \text{and} \qquad \frac{|z-a|}{\rho} < 1 .$$

The key point is writing the Cauchy kernel $\frac{1}{\zeta-z}$ as $$\frac{1}{\zeta-z} = \frac{1}{\zeta-a} \frac{\zeta-a}{\zeta-z} ,$$ and then using the geometric series.

Not only have we proved that $f$ has a power series, but we computed that the radius of convergence is at least $R$, where $R$ is the maximum $R$ such that $\Delta_R(a) \subset U$. We also obtained a formula for the coefficients $$c_k = \frac{1}{2\pi i} \int_{\gamma} \frac{f(\zeta )}{(\zeta -a)^{k+1}} d \zeta .$$

For a set $K$, denote the supremum norm: $$||f||_K \overset{\text{def}}{=} \sup_{z \in K} |f(z)| .$$ By a brute force estimation, we obtain the very useful Cauchy estimates: $$|c_k| = \left| \frac{1}{2\pi i} \int_{\gamma} \frac{f(\zeta )}{(\zeta -a)^{k+1}}d\zeta \right|\leq\frac{1}{2\pi }\int_{\gamma}\frac{||f||_{y}}{\rho^{k+1}}|d\zeta |=\frac{||f||_{y}}{\rho^{k}}.$$

We differentiate Cauchy’s formula $k$ times (using the Wirtinger $\frac{\partial}{\partial z}$ operator),

$$f^{(k)}(z)=\frac{\partial^{k}f}{\partial z^{k}}(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{k! f(\zeta )}{(\zeta -z)^{k+1}}d\zeta ,$$ and therefore $$k! \, c_k = f^{(k)}(a) = \frac{\partial^k f}{\partial z^k}(a) .$$ Hence, we can control derivatives of $f$ by the size of the function: $$|f^{(k)}(a)| = \left|\frac{\partial^k f}{\partial z^k}(a)\right| \leq \frac{k! ||f||_{\gamma}}{\rho^{k}} .$$ This estimate is one of the key properties of holomorphic functions, and the reason why the correct topology for the set of holomorphic functions is the same as the topology for continuous functions. Consequently, obstructions to solving problems in complex analysis are often topological in character.

For further review of one-variable results, see Appendix B.