Analytic Functions: Limits, Continuity, Polynomials, Rational Functions, Power Series, Maclaurin Series, Uniform Convergence, and Abel's Limit Theorem

Introduction to Analytic Functions

In complex analysis, the concept of an analytic function is central. An analytic function is a function that is differentiable at every point in some open set in the complex plane. This is a much stronger condition than mere differentiability in real calculus. If a function is analytic in an open set $D$, it means it is differentiable at every point within $D$.

Limits of Complex Functions

The concept of a limit for a complex function $f(z)$ as $z$ approaches a point $z_0$ is similar to that in real calculus. We say that the limit of $f(z)$ as $z$ approaches $z_0$ is $L$, denoted by $\lim_{z \to z_0} f(z) = L$, if for every $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |z - z_0| < \delta$, then $|f(z) - L| < \epsilon$.

The key difference is that $z$ can approach $z_0$ from any direction in the complex plane. For the limit to exist, the function must approach the same value $L$ regardless of the path taken by $z$ to reach $z_0$.

Continuity of Complex Functions

A complex function $f(z)$ is said to be continuous at a point $z_0$ if three conditions are met:

• $f(z_0)$ is defined.
• $\lim_{z \to z_0} f(z)$ exists.
• $\lim_{z \to z_0} f(z) = f(z_0)$.

If $f(z)$ is continuous at every point in an open set $D$, then $f(z)$ is continuous in $D$. Properties of limits and continuity in real calculus extend to complex functions, such as the sum, difference, product, and quotient (where the denominator is non-zero) of continuous functions are also continuous.

Differentiability and Analyticity

A complex function $f(z)$ is differentiable at a point $z_0$ if the limit

$f'(z_0) = \lim_{\Delta z \to 0} \frac{f(z_0 + \Delta z) - f(z_0)}{\Delta z}$

exists. If this limit exists, $f'(z_0)$ is the derivative of $f$ at $z_0$.

A function $f(z)$ is analytic in an open region $D$ if it is differentiable at every point in $D$. If a function is analytic in $D$, it is also continuous in $D$. The term "holomorphic" is often used interchangeably with "analytic".

Cauchy-Riemann Equations

A fundamental tool for determining if a complex function is analytic is the Cauchy-Riemann equations. Let $f(z) = u(x, y) + iv(x, y)$, where $z = x + iy$ and $u$ and $v$ are real-valued functions of two real variables $x$ and $y$. The function $f(z)$ is differentiable at $z_0 = x_0 + iy_0$ if and only if the first partial derivatives of $u$ and $v$ exist in a neighborhood of $(x_0, y_0)$, are continuous at $(x_0, y_0)$, and satisfy the Cauchy-Riemann equations at $(x_0, y_0)$:

$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$

If these equations hold and the partial derivatives are continuous in an open set $D$, then $f(z)$ is analytic in $D$.

The derivative of an analytic function can be expressed in terms of these partial derivatives:

$f'(z) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}$ or $f'(z) = \frac{\partial v}{\partial y} - i \frac{\partial u}{\partial y}$

Analyticity of Polynomials and Rational Functions

Polynomials in $z$ are always analytic functions over the entire complex plane $\mathbb{C}$. A general polynomial is of the form $P(z) = a_n z^n + a_{n-1} z^{n-1} + \dots + a_1 z + a_0$, where $a_i$ are complex coefficients. Since $z$ is analytic and powers of $z$ ($z^k$) are analytic, and sums and products of analytic functions are analytic, any polynomial is analytic everywhere.

Rational functions are functions of the form $R(z) = \frac{P(z)}{Q(z)}$, where $P(z)$ and $Q(z)$ are polynomials. A rational function is analytic everywhere except at the points where the denominator $Q(z)$ is zero. These points are the poles of the rational function.

Power Series

A power series is an infinite series of the form

$\sum_{n=0}^{\infty} a_n (z - z_0)^n = a_0 + a_1 (z - z_0) + a_2 (z - z_0)^2 + \dots$

where $a_n$ are complex coefficients and $z_0$ is a complex number called the center of the series.

For any given power series, there is a radius of convergence, $R$, which is a non-negative real number or infinity. The series converges absolutely for $|z - z_0| < R$ and diverges for $|z - z_0| > R$. The behavior at $|z - z_0| = R$ depends on the specific series.

A key theorem states that a function defined by a power series is analytic within its circle of convergence. Specifically, if $f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n$ with radius of convergence $R > 0$, then $f(z)$ is analytic in the open disk $|z - z_0| < R$. Furthermore, its derivative $f'(z)$ can be obtained by differentiating the series term by term:

$f'(z) = \sum_{n=1}^{\infty} n a_n (z - z_0)^{n-1}$ for $|z - z_0| < R$.

Maclaurin Series

A Maclaurin series is a special case of a Taylor series where the center $z_0$ is the origin ($z_0 = 0$). If a function $f(z)$ is analytic in a disk centered at the origin with radius $R$, then it can be represented by its Maclaurin series:

$f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n = f(0) + f'(0)z + \frac{f''(0)}{2!}z^2 + \frac{f'''(0)}{3!}z^3 + \dots$

This series converges to $f(z)$ for $|z| < R$. The coefficients of the Maclaurin series are determined by the values of the function and its derivatives at $z=0$.

Common Maclaurin series expansions include:

• $e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!} = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \dots$ (converges for all $z$)
• $\sin(z) = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!} = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots$ (converges for all $z$)
• $\cos(z) = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!} = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \dots$ (converges for all $z$)
• $\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n = 1 + z + z^2 + z^3 + \dots$ (converges for $|z| < 1$)

Uniform Convergence

Uniform convergence is a stronger form of convergence for sequences and series of functions. For a sequence of functions $f_n(z)$ converging to $f(z)$ on a domain $D$, the convergence is uniform if for every $\epsilon > 0$, there exists a single $N$ (independent of $z$) such that for all $n > N$, $|f_n(z) - f(z)| < \epsilon$ for all $z \in D$.

For a series of functions $\sum_{n=1}^{\infty} f_n(z)$, uniform convergence means that the sequence of partial sums $S_N(z) = \sum_{n=1}^{N} f_n(z)$ converges uniformly. The Weierstrass M-test is a useful tool to establish uniform convergence: If there exists a sequence of positive numbers $M_n$ such that $|f_n(z)| \leq M_n$ for all $z$ in the domain, and $\sum_{n=1}^{\infty} M_n$ converges, then the series $\sum_{n=1}^{\infty} f_n(z)$ converges uniformly on that domain.

Uniform convergence is crucial because it allows us to interchange limits and integrals, and it preserves analyticity. If a sequence of analytic functions $f_n(z)$ converges uniformly to $f(z)$ on a domain $D$, then the limit function $f(z)$ is also analytic in $D$. Furthermore, the derivatives of the functions also converge uniformly on compact subsets of $D$:

$f_n'(z) \to f'(z)$ uniformly on compact subsets of $D$.

Abel's Limit Theorem

Abel's Limit Theorem relates the limit of a function defined by a power series at the boundary of its disk of convergence to the convergence of the series itself.

Let $f(z) = \sum_{n=0}^{\infty} a_n z^n$ be a power series with radius of convergence $R=1$. Suppose the series converges at a point $z=1$. If $f(z)$ is continuous on the closed disk $|z| \leq 1$, then the limit of $f(z)$ as $z$ approaches 1 from within the disk exists and is equal to the sum of the series at $z=1$.

$\lim_{z \to 1, |z|<1} f(z) = \sum_{n=0}^{\infty} a_n$

More generally, if the series converges at a point $z_1$ on the circle of convergence $|z|=R$, and if $f(z)$ is continuous on the line segment connecting $0$ to $z_1$ (including $z_1$), then

$\lim_{r \to 1^-} f(r z_1) = \sum_{n=0}^{\infty} a_n (z_1)^n$

This theorem is particularly useful for evaluating series sums at the boundary where direct summation might be difficult. It connects the analytic behavior inside the disk to the behavior at the boundary.

Example: Polynomials

Consider the polynomial $P(z) = z^2 + 3z + 2$. This function is analytic everywhere in the complex plane $\mathbb{C}$. Its derivatives are $P'(z) = 2z + 3$, $P''(z) = 2$, and $P^{(n)}(z) = 0$ for $n \geq 3$. The Cauchy-Riemann equations are satisfied: if $f(z) = (x+iy)^2 + 3(x+iy) + 2 = (x^2 - y^2 + 2ixy) + (3x + 3iy) + 2 = (x^2 - y^2 + 3x + 2) + i(2xy + 3y)$, then $u(x,y) = x^2 - y^2 + 3x + 2$ and $v(x,y) = 2xy + 3y$. We have $\frac{\partial u}{\partial x} = 2x + 3$, $\frac{\partial v}{\partial y} = 2x + 3$, $\frac{\partial u}{\partial y} = -2y$, and $\frac{\partial v}{\partial x} = 2y$. This does not seem to satisfy the CR equations directly. Let's re-evaluate $P(z)$ in terms of $x$ and $y$.

Let $z = x+iy$. Then $z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy$. $P(z) = (x^2 - y^2 + 2ixy) + 3(x+iy) + 2 = (x^2 - y^2 + 3x + 2) + i(2xy + 3y)$. So, $u(x, y) = x^2 - y^2 + 3x + 2$ and $v(x, y) = 2xy + 3y$. $\frac{\partial u}{\partial x} = 2x + 3$ $\frac{\partial v}{\partial y} = 2x + 3$ $\frac{\partial u}{\partial y} = -2y$ $\frac{\partial v}{\partial x} = 2y$ The Cauchy-Riemann equations are $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$. Here, $2x+3 = 2x+3$ (satisfied). And $-2y = -(2y)$ (satisfied). Since the partial derivatives are continuous and satisfy the Cauchy-Riemann equations everywhere, $P(z)$ is analytic everywhere.

Example: Rational Functions

Consider $f(z) = \frac{1}{z}$. This function is analytic everywhere except at $z=0$. Let $z = x+iy$. Then $\frac{1}{z} = \frac{1}{x+iy} = \frac{x-iy}{(x+iy)(x-iy)} = \frac{x-iy}{x^2+y^2} = \frac{x}{x^2+y^2} + i \frac{-y}{x^2+y^2}$. So, $u(x, y) = \frac{x}{x^2+y^2}$ and $v(x, y) = \frac{-y}{x^2+y^2}$. We compute the partial derivatives: $\frac{\partial u}{\partial x} = \frac{(x^2+y^2)(1) - x(2x)}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}$ $\frac{\partial v}{\partial y} = \frac{(x^2+y^2)(-1) - (-y)(2y)}{(x^2+y^2)^2} = \frac{-x^2-y^2+2y^2}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}$ $\frac{\partial u}{\partial y} = \frac{(x^2+y^2)(0) - x(2y)}{(x^2+y^2)^2} = \frac{-2xy}{(x^2+y^2)^2}$ $\frac{\partial v}{\partial x} = \frac{(x^2+y^2)(0) - (-y)(2x)}{(x^2+y^2)^2} = \frac{2xy}{(x^2+y^2)^2}$ Checking the Cauchy-Riemann equations: $\frac{\partial u}{\partial x} = \frac{y^2-x^2}{(x^2+y^2)^2}$ and $\frac{\partial v}{\partial y} = \frac{y^2-x^2}{(x^2+y^2)^2}$. They are equal. $\frac{\partial u}{\partial y} = \frac{-2xy}{(x^2+y^2)^2}$ and $-\frac{\partial v}{\partial x} = -\frac{2xy}{(x^2+y^2)^2}$. They are equal. Since the partial derivatives are continuous for all $(x, y) \neq (0, 0)$ and satisfy the Cauchy-Riemann equations, $f(z) = \frac{1}{z}$ is analytic everywhere except at $z=0$.

Example: Uniform Convergence and Abel's Theorem

Consider the power series $\sum_{n=1}^{\infty} \frac{z^n}{n^2}$. This series has radius of convergence $R=1$. Let's check convergence at $z=1$. The series becomes $\sum_{n=1}^{\infty} \frac{1}{n^2}$, which is a convergent p-series ($p=2 > 1$). Now consider the function $f(z) = \sum_{n=1}^{\infty} \frac{z^n}{n^2}$ for $|z| < 1$. The function $f(z)$ is analytic inside the unit disk. Abel's Limit Theorem states that if $f(z)$ is continuous on the closed disk $|z| \leq 1$, then $\lim_{z \to 1, |z|<1} f(z) = \sum_{n=1}^{\infty} \frac{1}{n^2}$. The series $\sum_{n=1}^{\infty} \frac{z^n}{n^2}$ converges uniformly on the closed unit disk $|z| \leq 1$ by the Weierstrass M-test, since $|\frac{z^n}{n^2}| \leq \frac{1}{n^2}$ and $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges. Uniform convergence on the closed disk implies continuity on the closed disk. Thus, Abel's theorem applies.

The sum of this series is known to be $\frac{\pi^2}{6}$. So, $\lim_{z \to 1, |z|<1} \sum_{n=1}^{\infty} \frac{z^n}{n^2} = \frac{\pi^2}{6}$.

Summary of Key Concepts

Understanding analytic functions requires grasping several interconnected ideas:

• Analyticity: Differentiability in an open set.
• Cauchy-Riemann Equations: Necessary and sufficient conditions for analyticity.
• Polynomials and Rational Functions: Examples of analytic functions with specific points of non-analyticity for rational functions.
• Power Series: Represent analytic functions within their radius of convergence and can be differentiated term-by-term.
• Uniform Convergence: Ensures that the limit of analytic functions is analytic and allows for term-by-term differentiation of series.
• Abel's Limit Theorem: Connects the sum of a power series at a boundary point (if convergent) to the limit of the function at that point, provided continuity.