As long as algebra is taught in school, there will be prayer in school, Cokie Roberts

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Recall: Relationship Between Complex Differentiability and the Cauchy-Riemann Equations

Necessary Condition: Complex Differentiability Implies Cauchy-Riemann Equations

If a function f(z) = u(x, y) + iv(x, y) is complex-differentiable at a point z₀ =x₀ +iy₀, then its real and imaginary parts, u(x,y) and v(x,y), must satisfy the Cauchy-Riemann equations at that point: $\frac{\partial u}{\partial x}(x_0, y_0) = \frac{\partial v}{\partial y}(x_0, y_0), \frac{\partial u}{\partial y}(x_0, y_0) = -\frac{\partial v}{\partial x}(x_0, y_0)$.

Sufficient Condition: Cauchy-Riemann Equations (with Continuous Partial Derivatives) Imply Complex Differentiability

Proposition. Let f(z) = u(x, y) + iv(x, y) for z = x + iy ∈ D where D ⊆ ℂ is an open set. Assume that u and v have continuous first partial derivatives throughout D and they satisfy the Cauchy-Riemann equations at a point z ∈ D. Then, f'(z) exists, i.e., f is differentiable at z.

Analytic functions

Definition. A complex function f is said to be analytic (holomorphic) at a point z₀ if it satisfies any of the following equivalent conditions:

Differentiable in a neighborhood. There exists an open neighborhood U of z₀ such that f is complex differentiable at every point in U. Complex differentiability means that the derivative f′(z₀) exists for all z₀ ∈ U in the complex sense, f′(z₀) = $\lim_{h \to 0} \frac{f(z₀+h)-f(z₀)}{h}$.
Here the limit must be the same regardless of the direction from which h approaches zero in the complex plane.
Open disc criterion. There exists some radius r > 0 such that f is complex differentiable at every point in the open disc B(z₀ ; r) = {z: |z - z₀| < r}. This is a specific case of Condition 1, as an open disc is a specific type of open neighborhood around z₀.
Power series expansion. The function can be locally expressed as a convergent power series:$f(z) = \sum_{n = 0}^\infty a_n(z -z₀)^n$ in some neighborhood of z₀. This series converges absolutely and uniformly on compact subsets within its disk of convergence.

Recall.

A series is absolutely convergent if: $\sum_{n = 0}^\infty |a_n(z - z_0)^n|$ converges. This means the terms shrink fast enough that even their magnitudes add up to a finite value. It’s a strong form of convergence—if a series converges absolutely, it also converges normally.
Uniform convergence means the series converges at the same rate across the entire subset —not just pointwise. It’s like everyone in the race finishing together, no stragglers.
For any power series $\sum_{n = 0}^\infty a_n(z - z_0)^n$, there exists a radius of convergence R ≥ 0 such that: (1) The series converges absolutely for all z with ∣z − z₀∣ < R. The series diverges for all z with ∣z − z₀∣ > R. This defines the disk of convergence: ∣z − z₀∣ < R.
A compact subset is any closed and bounded region inside that disk.

While it is a straightforward consequence that a function defined by a convergent power series is complex differentiable, the converse is a profound and remarkable result. This equivalence underscores the extraordinary strength of complex differentiability. It means that the seemingly simple condition of being complex differentiable once in an open set automatically guarantees that the function is infinitely differentiable and can be locally expressed as a power series.

A function is defined as holomorphic (analytic) on an open set U ⊆ ℂ if it is complex differentiable at every point in U.

The concept of analyticity is a cornerstone of complex analysis, representing a significant strengthening of the notion of differentiability. While a function of a real variable can be differentiable at a single point without being differentiable in any neighborhood of that point, a function of a complex variable that is analytic at a point must be differentiable in an open set containing that point, this is what gives analytic functions their remarkable properties.

Definition. A function f which is analytic (holomorphic) on the entire complex plane ℂ is called an entire function.

Properties

Analytic (holomorphic) functions enjoy many powerful properties and obey strict rules. Below are some of the most important properties, assuming f and g are analytic in some domain (open set) D:

Infinite differentiability & Power series expansion. If a function is analytic at a point, it is guaranteed that its derivatives of all orders exist at that point. Unlike real functions, where a single derivative does not guarantee higher-order derivatives, the existence of a complex derivative in a neighborhood immediately implies an infinite hierarchy of derivatives. More strongly, f can be expanded in a convergent power series around $z_0$
Cauchy-Riemann equations. For a complex function f(z)=u(x,y)+iv(x,y), where u and v are real-valued functions of x and y, to be analytic in a domain D, it is both necessary and sufficient that u and v possess continuous first-order partial derivatives satisfying the equations: $\frac{du}{dx} = \frac{dv}{dy} \text{ and } \frac{du}{dy} = -\frac{dv}{dx}$ within D.
Closure under addition, subtraction, and multiplication.Suppose f and g are analytic at some point a ∈ ℂ, then f + g, f - g, f · g are also analytic at a. In words, the properties of analyticity are preserved under the standard arithmetic operations of addition, subtraction, and multiplication.

Proof.

Suppose f and g are two functions that are analytic at a point a ∈ ℂ. Then, the functions f + g, f − g, and f⋅g are also analytic at the point a. This property follows directly from the definition of analyticity and the corresponding rules for differentiation.

If f and g are differentiable in an open neighborhood of a, then their sum, difference, and product are also differentiable in that same neighborhood, with derivatives given by the familiar sum, difference, and product rules from calculus: h(x) = f(x) ± g(x), then h’(x) = f’(x) ± g’(x), (f(x) * g(x))’ = f’(x) * g(x) + f(x) * g’(x).

Closure under division. If f and g are analytic in D and g(a) ≠ 0, then the quotient function f/g is also analytic at the point a. This is because if g is analytic at a and g(a) ≠ 0, then by continuity there exists some open neighborhood around a where $g(z)\neq 0$. In this neighborhood, the quotient f(z)/g(z) is well-defined and differentiable (f/g is analytic), with its derivative given by the quotient rule: (f/g)′ = $\frac{f′(z)g(z)−f(z)g′(z)}{g(z)^2}$ which exists as long as $g(z)\neq0$.
Polynomials and rational functions. Since constant functions and the identity function f(z) = z are entire, it follows that all polynomial functions, which are sums and products of these, are also entire. Similarly, rational functions, which are quotients of polynomials, are analytic at every point in their domain (i.e., where the denominator is non-zero).

A rational function $r(z) = \frac{p(z)}{q(z)}$ (with $p$ and $q$ polynomials) is holomorphic everywhere except where $q(z)=0$ (its poles). For example, $f(z)=\frac{1}{z^2+1}$ is analytic for all $z$ except at the points $z=i$ and $z=-i$ (the zeros of the denominator), which are poles.

Identity Theorem.

Formulation 1. Zeros of an Analytic Function. Let D ⊂ ℂ be a domain (an open connected set), and f : D → ℂ be an analytic function. If there exists an infinite sequence {zk} ⊂ D, such that:

f(zₖ) = 0, ∀k ∈ ℕ
zₖ → z₀ ∈ D (i.e., the sequence converges to a point in D), then f(z) = 0 for all z ∈ D. If an analytic function is zero on a set with an accumulation point in D, it must be identically zero.

Formulation 2. Equality of Two Analytic Functions.

Let f(z) and g(z) be analytic in a domain D. If the set of points where f(z) = g(z) has a limit point in D (i.e., there exists a convergent sequence zₖ →z₀ ∈ D with f(zₖ) = g(zₖ)), then f(z) = g(z) for all z ∈ D. Two analytic functions that agree on a set with a limit point must be identical everywhere in D.

Example: Suppose f is analytic in ℂ (entire) and f(1/n) = 0 for all n ∈ ℕ. Since $\frac{1}{n} \to 0$ and 0 is in the domain, the Identity Theorem implies f(z) = 0 everywhere.

Proof (Formulation 1): Since f is analytic at z₀, it has a power series expansion in some neighborhood Bᵣ(z₀) ⊂ D: $f(z) = \sum_{n = 0}^∞ a_n(z - z_0)^n$.

Our goal is to show that all coefficients aₙ = 0, forcing f ≡ 0 near z₀.

Since f(zₖ) = 0 and zₖ → z₀, continuity of f implies, f(z₀) = $\lim_{k \to \infty} f(z_k) = 0$. But f(z₀) = a₀, so a₀ = 0.

Inductive argument for higher coefficients. Assume $a_0 = a_1 = \dots = a_{m-1} = 0$. Then the power series becomes: $f(z) = \sum_{n = m}^∞ a_n(z - z_0)^n = (z-z_0)^m\sum_{n = 0}^∞ a_{n+m}(z - z_0)^n$

Define g(z) = $\sum_{n = 0}^∞ a_{n+m}(z - z_0)^n$, which is analytic in Bᵣ(z₀)

Since f(zₖ) = 0 and zₖ ≠ z₀ for large k, we have 0 = f(zₖ) = (zₖ - z₀)ᵐg(zₖ) ⇒ g(zₖ) = 0

Notice that $f(z)$ can be factored as $f(z) = (z - z_0)^mg(z)$. Because $f(z_k) = 0$ for all large $k$ (with $z_k \neq z_0$ for sufficiently large $k$ since $z_k \to z_0$ but $z_k \neq z_0$), we can cancel the factor $(z_k - z_0)^m$ to conclude $g(z_k) = 0$ for those $k$. In other words, $z_k$ (for large $k$) are zeros of $g(z)$ as well.

Taking k→∞, continuity of g gives g(z₀) = aₘ = 0. By induction, aₙ = 0 for all n, so f ≡ 0 in Bᵣ(z₀).

Extend to the entire domain D. Let A = {z ∈ D | f ≡ 0 in a neighborhood of z}. A is open (by definition), non-empty (z₀ ∈ A), and closed (if wₜ ∈ A and wₜ → w, then f≡0 near w by repeating the argument above).

Since D is connected, the only non-empty closed subset is D itself. Thus, A = D, meaning f≡0 everywhere in D.

Corollary (Equality of Two Analytic Functions) If f and g are analytic in D and agree on a set with a limit point in D, then h(z) = f(z) − g(z) is analytic and vanishes on that set. By the Identity Theorem, h ≡ 0, so f ≡ g.

Closure under composition. The composition of analytic functions is analytic. if f is an analytic function of z on a domain D, and g is an analytic function of w on a domain that contains the image f(D), then the composite function h(z)=g(f(z)) is analytic on D. The derivative of the composite function is given by the familiar chain rule: h′(z) = g′(f(z))⋅f′(z), e.g. $e^z, cos(z^2 + 1)$ are entire functions because they are compositions of entire functions.

Proof:

We proceed by considering two cases for the behavior of f(z) near z₀.

Case 1. f(z) ≠ f(z₀) for all z in a punctured neighborhood of z₀.

In this scenario, for z sufficiently close to z₀ (but z ≠ z₀), the term f(z) − f(z₀) is non-zero. This allows for a strategic manipulation of the difference quotient by multiplying and dividing by f(z)−f(z₀):

$ \frac{g(f(z)) - g(f(z_0))}{z - z_0} = \frac{g(f(z)) - g(f(z_0))}{f(z) - f(z_0)} \cdot \frac{f(z) - f(z_0)}{z - z_0}$

Now, taking the limit as z→z₀. For the first factor, we know that f(z) → f(z₀). Let w = f(z). Then the limit of the first factor becomes $\frac{g(w) - g(w_0)}{w - w_0}$. This is precisely the definition of the derivative of g at w₀, which is g′(w₀) = g′(f(z₀)), since g is analytic at w₀. The second factor directly corresponds to the definition of the derivative of f at z₀, which is f′(z₀), since f is analytic at z₀.

Therefore, in this case, the limit exists and is given by the product: $\lim_{z \to z_0} \frac{g(f(z)) - g(f(z_0))}{f(z) - f(z_0)} \cdot \frac{f(z) - f(z_0)}{z - z_0} = g'(f(z_0))f'(z_0)$.

Case 2. f(z) = f(z₀) for infinitely many z in any neighborhood of z₀.

This situation implies that z₀ is an accumulation point for the set of zeros of the function k(z) = f(z) - f(z₀). Since f is an analytic function, k(z) is also analytic (as the difference of two analytic functions). The Identity Theorem states that if an analytic function has an accumulation point of zeros within its domain, then the function must be identically zero throughout the connected component containing that accumulation point.

Consequently, if k(z₀) = 0 ↭ f(z) = f(z₀) for infinitely many points accumulating at z₀, then f(z) must be identically equal to f(z₀) for all z in some open neighborhood of z₀. If f(z) is constant in a neighborhood of z₀, its derivative f’(z) must be 0. In this scenario, for z ≠ z₀ within this neighborhood h(z) -h(z₀) = g(f(z)) - g(f(z₀)) = g(f(z₀)) - g(f(z₀)) = 0. Thus, the difference quotient becomes $\frac{0}{z-z_0} = 0$. Taking the limit as z → 0, we find that h’(z₀) = 0. Crucially, the formula derived in Case 1, $g'(f(z_0))f'(z_0)$ also yields $g'(f(z_0))·0 = 0$, demonstrating consistency across both cases.

Analytic inverse functions. If a function f(z) is analytic and one-to-one (injective) on a domain D, and its derivative f′(z) is never zero on D (i.e. $f'(z)\neq 0$ for all $z\in D$), then its inverse function f⁻¹ (w) is also analytic on the image f(D). The derivative of the inverse function is given by the formula: $\frac{d}{dw}f^{-1}(w) = \frac{1}{f'(f^{-1}(w))}$.

For example, $f(z)=z^3$ is entire and one-to-one on the domain D = {z ∈ ℂ | z ≠ 0 and $|arg(z)| < \frac{\pi}{3}$} (the cube roots of unity are 1, w = $e^{2\pi i/3}$, w² = $e^{4\pi i/3}$ —are spaced $120^\circ$ apart). If our domain included more than $\frac{2\pi}{3}$ of angular width, we could have multiple z values such that z³ lands on the same point. We must exclude z = 0 because f’(0) = 0, violating our key condition. Since f is a polynomial, so it’s analytic everywhere, $f'(z)=3z^2$ is never 0 on that domain, the inverse function $f^{-1}(w)=\sqrt[3]{w}$ is analytic on $f(\text{domain})$, $\frac{d}{dw}f^{-1}(w) = \frac{1}{3}\sqrt[3]{w^2}$

These closure properties make the class of analytic functions a very robust and well-behaved collection of objects to study.

Examples of Analytic Functions

Constant and identity functions: The simplest examples are constant functions f(z) = c (with c a constant complex number) and the identity function f(z) = z. Both are entire functions (analytic everywhere on $\mathbb{C}$). The derivative of a constant is 0, and the derivative of the identity function f(z) = z is 1, which exist everywhere, so they meet the criteria of analyticity on all of $\mathbb{C}$ and serve as building blocks for polynomials and more complicated analytic functions.
Polynomial functions are entire functions: Every polynomial $p(z) = a_n z^n + a_{n-1}z^{n-1} + \cdots + a_1 z + a_0$ is an entire function. For instance, $f(z)=z^2+3z+2, z^5 - 2z + 7$ is analytic for all $z \in \mathbb{C}$. This follows because polynomials are sums and products of the identity function and constants, and these operations preserve analyticity.

$f(z)=z^2+3z+2 = (x+iy)^2+3(x+iy)+2 = (x²-y² + 3x +2) + i(2xy + 3y)$. We compute the partial derivatives $\frac{\partial u}{\partial x} = 2x+3, \frac{\partial u}{\partial y} = -2y, \frac{\partial v}{\partial x} = 2y, \frac{\partial v}{\partial y}=2x+3$

Since the EC are satisfied and the partial derivatives are continuous everywhere, the function f(z) is analytic everywhere in the complex plane, i.e., it is an entire function.

The complex exponential function eᶻ and the complex trigonometric functions sin(z) and cos(z) are entire functions. In fact, these functions can be defined by their power series which converge for all $z$: $e^z = \displaystyle\sum_{n=0}^{\infty} \frac{z^n}{n!}, \cos z = \displaystyle\sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{(2n)!}, \sin z = \displaystyle\sum_{n=0}^{\infty} (-1)^n \frac{z^{2n+1}}{(2n+1)!}$.
Rational functions. Any rational function $R(z) = \frac{p(z)}{q(z)}$ (with p, q polynomials) is analytic on its domain (the set ${z: q(z)\neq 0}$).

For example, $h(z) = \frac{1}{z² + 1}$. The inner function is f(z) = z² + 1. This is a polynomial, and thus an entire function (analytic on C). The outer function if g(w) = $\frac{1}{w}$. This function is analytic for all w ≠ 0. For the composite function h(z) = g(f(z)) to be well-defined and analytic, the output of f(z) must not fall into the singularity of g(w). That is, f(z) ≠ 0 ↭ z² + 1 ≠ 0 ↭ The solutions are z ≠ i and z ≠ −i. Therefore, h(z) is analytic on the domain C∖{i,−i}. h′(z) = g′(f(z))·f′(z) = $-\frac{1}{(z² + 1)^2}·2z = -\frac{2z}{(z² + 1)^2}$

$z=\pm i$ are poles (isolated singularities) of h; everywhere else, h has well-defined complex derivatives of all orders. This illustrates how rational functions, though not entire, are locally analytic wherever they are defined.

Compositions of entire functions: If we take combinations of entire functions, the result is entire, e.g., h(z) = $e^{z²}$ is entire. Both the inner f(z) = z² and outer g(w) = eʷ functions are entire functions (a polynomial and the exponential functions are analytic on ℂ). The range of f and g are the entire complex plane ℂ. Since f(ℂ) ⊆ domain(g) = ℂ, the domain compatibility condition is met. Therefore, by the composition theorem, h(z) = g(f(z)) = $e^{z²}$ is an entire function. The derivative is obtained by the chain rule: h′(z) = g′(f(z))·f′(z) = $e^{z²}·2z = 2ze^{z²}$
h(z) = cos(z² + 1) is entire. Both the inner f(z) = z² + 1 and outer g(w) = cos(w) functions are entire functions (a polynomial and the cosine functions are analytic on ℂ). The range of f and g are the entire complex plane ℂ. Since f(ℂ) ⊆ domain(g) = ℂ, the domain compatibility condition is met. Therefore, by the composition theorem, h(z) = g(f(z)) = cos(z² + 1) is an entire function. Differentiating, h′(z) = g′(f(z))·f′(z) = -sin(z² + 1)·2z = -2zsin(z² + 1).
The complex logarithm function $\log (z)$ cannot be defined as a single-valued analytic function on all of $\mathbb{C}$ (since it’s multi-valued around 0, requiring a branch cut). However, if we restrict its domain to exclude a ray from the origin (for example, the principal branch excludes the negative real axis), then $\log (z)$ becomes analytic on that cut-plane, $\mathbb{C}\setminus\\{x\le0 : x\in\mathbb{R}\\}.$

The derivative of any analytic branch of the logarithm can be derived by differentiating the equation $e^{log(z)} = z$ using the chain rule. We have $e^{log(z)}·\frac{d}{dz}log(z) = 1$, which implies $\frac{d}{dz}log(z) = \frac{1}{e^{log(z)}} = \frac{1}{z}$.

h(z) = log(z² + 1). The inner function is f(z) = z² + 1. This is a polynomial, and thus an entire function (analytic on C). The outer function is g(w) = log(w).

For $h(z)$ to be well-defined and analytic, we must choose a branch for $\log$ and ensure $f(z)$ does not hit the branch cut or the singularity of $\log$. Suppose we use the principal branch of $\log$ (cut along $(-\infty,0]$ on the real axis) The condition is that z² + 1 must not lie on the non-positive real axis.

Let z = x + iy ⇒ z² + 1 = (x² - y² + 1) + i2xy.

z² + 1 lie on the non-positive real axis ↭ 2xy = 0 (x = 0 or y = 0), x² - y² + 1 ≤ 0

y = 0, x² - y² + 1 ≤ 0 ↭ x² + 1 ≤ 0 ⊥
x = 0 (purely imaginary z), x² - y² + 1 ≤ 0 ↭ -y² + 1 ≤ 0 ↭ y² ≥ 1 ↭ y ≥ 1 or y ≤ −1.

Thus, h(z) is analytic on {$z \in \mathbb{C}: z \neq i y \text{ for } |y| \ge 1$}. On this domain, $h(z)$ has a well-defined derivative given by the chain rule: h′(z) = g′(f(z))·f′(z) = $\frac{1}{z² + 1}·2z = \frac{2z}{z² + 1}$

Counterexamples (Non-Analytic Functions)

Not every complex function is analytic. In fact, complex-differentiability is a very stringent condition, and many well-behaved real functions fail to be holomorphic.

$f(z) = |z|^2 = x^2 + y^2$ is not analytic (except at a single point). This function is real-differentiable everywhere and even has continuous partial derivatives, but it does not satisfy the Cauchy-Riemann equations except at the origin.

$u(x,y)=x^2+y^2$ and $v(x,y)=0$ give $\frac{\partial u}{\partial x} = 2x$, $\frac{\partial v}{\partial y} = 0$, which are only equal at x = 0; likewise $\frac{\partial u}{\partial y} = 2y$, $\frac{\partial v}{\partial x} = 0$, only equal at $y=0$. So the only point where CR equations hold is $(0,0)$, and f is complex-differentiable at $z=0$ (the limit definition yields 0 in all directions)

$f'(z) = lim_{h \to 0} \frac{f(z+h)-f(z)}{h} = lim_{h \to 0} \frac{|z+h|^2-|z|^2}{h} =[\text{Recall: } |z+h|^2 = (z + h)\overline{z + h} = (z + h)(\bar z + \bar h) = |z|^2 + z\bar h + \bar zh + |h|^2] = lim_{h \to 0} \frac{|z|^2 + z\bar h + \bar zh + |h|^2-|z|^2}{h} = lim_{h \to 0} z·\frac{\bar h}{h} + \bar z + \frac{h\bar h}{h} = lim_{h \to 0} z·\frac{\bar h}{h} + \bar z$

If h = t ∈ ℝ, $lim_{h \to 0} z·\frac{\bar h}{h} + \bar z = z·1 + \bar z = 2\mathbb{R(z)}$
If h = it, t ∈ ℝ, $lim_{h \to 0} z·\frac{\bar h}{h} + \bar z = z·(-1) + \bar z = -2i\mathbb{I(z)}$. Note: $- z + \overline{z} = - (x + iy) + (x - iy) = -x - iy + x - iy = -2iy$.
Conclusion: These are not equal unless z = 0. ❌ So the limit does not exist for z ≠ 0.

|z|² is not complex differentiable in any neighborhood of 0. No matter how small a radius around 0 you take, there will be points where the derivative does not exist (actually, every other point). At z = 0, the function is “flat enough” for the complex derivative to exist. But elsewhere, it grows radially. Therefore $f(z)=|z|^2$ is not holomorphic at 0 (or anywhere else).

This example underscores that being differentiable at an isolated point (as $|z|^2$ is at 0) is not sufficient for analyticity; analyticity demands differentiability in a full neighborhood. In summary, $|z|^2$ is not analytic on any open set (it fails to meet the criteria of the definition), even though it is differentiable at the single point $0$.

The complex conjugate $z$ is a classic example of a function that is not analytic anywhere. $u(x,y)=x$, $v(x,y)=-y$, giving $\frac{\partial u}{\partial x} = 1$, $\frac{\partial v}{\partial y} = -1$, which immediately violates the Cauchy-Riemann equation $u_x = v_y$
Similarly, the functions $\Re(z) = x$ and $\Im(z) = y$ are not analytic (except in trivial cases), nor is $|z| = \sqrt{x^2+y^2}$.
Finding all solutions to the equation eᶻ = 2 + 2i.

First, express 2 + 2i in polar form. $r =|eᶻ| = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}, tan(\theta) = \frac{2}{2} = 1$. Since the point (2,2) is in the first quadrant, the principal argument is $\theta = \frac{\pi}{4}, \text{so } 2 + 2i = 2\sqrt{2}e^{iπ/4}$.
Now, we apply the general solution formula: $z = ln(2\sqrt{2}) + i(π/4 + 2kπ)$, for any integer k.
Simplifying this, we get $ln(2\sqrt{2}) = ln(2^{1+1/2}) = ln(2^{3/2}) = \frac{3}{2}ln(2)$
Thus, the solutions are: $z = \frac{3}{2}ln(2) + i(π/4 + 2kπ)$, for any integer k.

Analytic Functions in Complex Analysis