Index of a point with respect to a closed curve

The index of a point with respect to a closed curve is a fundamental concept in complex analysis that quantifies how many times a closed curve winds around a given point. It's also known as the winding number or the rotation number. This concept is crucial for understanding Cauchy's Integral Theorem and Cauchy's Integral Formula, and it forms the basis for many advanced theorems in the field.

Definition of the Index

Let $\gamma$ be a closed curve in the complex plane, and let $z_0$ be a point not on $\gamma$. The index of $z_0$ with respect to $\gamma$, denoted by $n(\gamma, z_0)$ or $Ind(\gamma, z_0)$, is defined as:

$$n(\gamma, z_0) = \frac{1}{2\pi i} \int_{\gamma} \frac{dz}{z - z_0}$$

If $\gamma$ is a piecewise smooth curve, the integral is interpreted as the sum of integrals over the smooth segments.

Geometric Interpretation

Geometrically, the index represents the total change in the argument of $(z - z_0)$ as $z$ traverses the curve $\gamma$, divided by $2\pi$. If we parameterize the curve $\gamma$ by $z(t)$ for $t \in [a, b]$, then:

$$\int_{\gamma} \frac{dz}{z - z_0} = \int_{a}^{b} \frac{z'(t)}{z(t) - z_0} dt$$

Consider the logarithm of $(z - z_0)$. Let $w = z - z_0$. As $z$ moves along $\gamma$, $w$ traces a curve starting and ending at the same point (since $\gamma$ is closed). Let $w(t) = r(t) e^{i\theta(t)}$. Then $\log(w(t)) = \log(r(t)) + i\theta(t)$.

The derivative of $\log(w(t))$ with respect to $t$ is:

$$\frac{d}{dt} \log(w(t)) = \frac{w'(t)}{w(t)} = \frac{z'(t)}{z(t) - z_0}$$

Integrating this from $a$ to $b$:

$$\int_{a}^{b} \frac{z'(t)}{z(t) - z_0} dt = \log(w(b)) - \log(w(a))$$

Since $\gamma$ is closed, $z(a) = z(b)$, so $w(a) = w(b)$. However, the logarithm is multi-valued. If $\log(w(b)) = \log(r(b)) + i(\theta(b) + 2\pi k_1)$ and $\log(w(a)) = \log(r(a)) + i(\theta(a) + 2\pi k_2)$, where $\theta(b)$ and $\theta(a)$ are the principal arguments, then:

$$\int_{\gamma} \frac{dz}{z - z_0} = (\log(r(b)) + i\theta(b) + 2\pi i k_1) - (\log(r(a)) + i\theta(a) + 2\pi i k_2)$$

Since $r(a) = r(b)$ and $\theta(a)$ and $\theta(b)$ differ by a multiple of $2\pi$, let $\theta(b) = \theta(a) + 2\pi N$ for some integer $N$.

Then the integral becomes $i(2\pi N + 2\pi i (k_1 - k_2))$. The imaginary part of the integral is $2\pi N$.

So, $\frac{1}{2\pi i} \int_{\gamma} \frac{dz}{z - z_0} = N$. This integer $N$ is the index. It represents the total change in the argument of $(z - z_0)$ divided by $2\pi$.

Properties of the Index

1. If $z_0$ is outside the region bounded by $\gamma$, then $n(\gamma, z_0) = 0$.
2. If $z_0$ is inside the region bounded by $\gamma$, and $\gamma$ is oriented counterclockwise, then $n(\gamma, z_0) = 1$.
3. If $z_0$ is inside the region bounded by $\gamma$, and $\gamma$ is oriented clockwise, then $n(\gamma, z_0) = -1$.
4. The index is an integer.
5. If $\gamma_1$ and $\gamma_2$ are two closed curves, and $z_0$ is not on either curve, then $n(\gamma_1 + \gamma_2, z_0) = n(\gamma_1, z_0) + n(\gamma_2, z_0)$. This means the index is additive for concatenated curves.
6. If $\gamma_1$ and $\gamma_2$ are homotopic with respect to $z_0$ (i.e., they can be continuously deformed into each other without passing through $z_0$), then $n(\gamma_1, z_0) = n(\gamma_2, z_0)$.
7. The function $f(z) = n(\gamma, z)$ is constant on each connected component of $\mathbb{C} \setminus \gamma$. Specifically, it is 0 on the unbounded component (the exterior).

Example

Consider a circle $\gamma$ defined by $|z| = R$ traversed counterclockwise. Let $z_0$ be a point.

Case 1: $z_0$ is inside the circle, i.e., $|z_0| < R$. We can parameterize $\gamma$ as $z(t) = R e^{it}$ for $t \in [0, 2\pi]$. Then $z'(t) = i R e^{it}$.

$$n(\gamma, z_0) = \frac{1}{2\pi i} \int_{0}^{2\pi} \frac{i R e^{it}}{R e^{it} - z_0} dt = \frac{1}{2\pi} \int_{0}^{2\pi} \frac{R e^{it}}{R e^{it} - z_0} dt$$

Let $w = z - z_0$. As $z$ traverses the circle, $w$ traverses a circle of radius $R$ centered at $-z_0$. The integral $\int_{\gamma} \frac{dz}{z-z_0}$ is the integral of $\frac{1}{w}$ around a circle centered at $-z_0$.

Alternatively, using the property that the function $n(\gamma, z)$ is constant on connected components: For $|z| = R$ traversed counterclockwise, $n(\gamma, z) = 1$ for $|z| < R$ and $n(\gamma, z) = 0$ for $|z| > R$. So, if $|z_0| < R$, $n(\gamma, z_0) = 1$.

Case 2: $z_0$ is outside the circle, i.e., $|z_0| > R$. Then $n(\gamma, z_0) = 0$.

Case 3: $z_0$ is on the circle, i.e., $|z_0| = R$. The definition of the index requires $z_0$ not to be on $\gamma$. The integral is improper in this case.

Local Properties of Analytic Functions

Analytic functions possess remarkable local properties that distinguish them from more general differentiable functions. These properties arise from the Cauchy-Riemann equations and the fact that analytic functions are infinitely differentiable and possess power series representations.

Definition of Analytic Function

A function $f(z)$ is said to be analytic in an open set $D$ if it is differentiable at every point $z$ in $D$. A function is analytic at a point $z_0$ if it is analytic in some neighborhood of $z_0$.

Key Local Properties

1. Infinite Differentiability: If $f(z)$ is analytic in a domain $D$, then $f'(z), f''(z), f'''(z), \ldots$ all exist for every $z \in D$. This is a powerful property. For real differentiable functions, the existence of the first derivative does not guarantee the existence of the second derivative.
2. Power Series Representation (Taylor's Theorem): If $f(z)$ is analytic in a domain $D$, then for any point $z_0 \in D$, $f(z)$ can be represented by a Taylor series expansion around $z_0$ that converges in some open disk centered at $z_0$ and contained within $D$.
3. Cauchy's Integral Formula: If $f(z)$ is analytic in a simply connected domain $D$, and $\gamma$ is a simple closed contour within $D$, and $z_0$ is any point inside $\gamma$, then: $$f(z_0) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(z)}{z - z_0} dz$$ This formula shows that the value of an analytic function at any interior point is completely determined by its values on the boundary.
4. Liouville's Theorem: A bounded entire function (analytic on the entire complex plane) must be constant. This is a global property but has local implications.
5. Maximum Modulus Principle: If $f(z)$ is analytic and non-constant in a domain $D$, then $|f(z)|$ has no local maximum in $D$. If $D$ is bounded and $f$ is continuous on its closure $\bar{D}$, then the maximum value of $|f(z)|$ must occur on the boundary of $D$.
6. Open Mapping Theorem: If $f(z)$ is analytic and non-constant in a domain $D$, then $f(D)$ is an open set. This means that analytic functions map open sets to open sets.
7. Zeros of Analytic Functions: The zeros of a non-constant analytic function are isolated. This means that if $f(z_0) = 0$ for some $z_0$ in a domain $D$, then there exists a neighborhood around $z_0$ where $f(z) \neq 0$ except at $z_0$ itself.

The Role of the Cauchy-Riemann Equations

The Cauchy-Riemann equations are a necessary condition for a function to be differentiable. Let $f(z) = u(x, y) + i v(x, y)$, where $z = x + iy$. The Cauchy-Riemann equations are:

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

If $f(z)$ is analytic in a domain $D$, then $u$ and $v$ are not only continuous but also possess continuous first partial derivatives that satisfy the Cauchy-Riemann equations. Furthermore, the second partial derivatives of $u$ and $v$ are also continuous, and they satisfy Laplace's equation:

$$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \quad \text{and} \quad \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0$$ This means the real and imaginary parts of an analytic function are harmonic functions.

Removable Singularities

Singularities are points where a complex function fails to be analytic. In complex analysis, we classify these singularities into three types: removable singularities, poles, and essential singularities. Understanding these classifications is vital for contour integration and series expansions.

Definition of a Singularity

A point $z_0$ is called an isolated singularity of a function $f(z)$ if $f(z)$ is analytic in some punctured disk $0 < |z - z_0| < R$ for some $R > 0$, but $f(z)$ is not analytic at $z_0$.

Removable Singularities

A singularity $z_0$ is called removable if there exists an analytic function $g(z)$ defined in a neighborhood of $z_0$ such that $f(z) = g(z)$ for all $z$ in the punctured neighborhood $0 < |z - z_0| < R$. In essence, the singularity can be "removed" by defining or redefining the function's value at $z_0$.

Conditions for a Removable Singularity

There are several equivalent conditions to identify a removable singularity $z_0$:

1. Limit Exists: $\lim_{z \to z_0} f(z)$ exists and is finite. If the limit exists, say $L$, then $z_0$ is removable, and the function can be made analytic by defining $f(z_0) = L$.
2. Boundedness: $f(z)$ is bounded in some punctured neighborhood of $z_0$, i.e., there exists a constant $M > 0$ such that $|f(z)| \leq M$ for all $z$ in $0 < |z - z_0| < R$. This is known as Riemann's Theorem on Removable Singularities.
3. Laurent Series: The Laurent series expansion of $f(z)$ around $z_0$ has no terms with negative powers of $(z - z_0)$. That is, the principal part of the Laurent series is zero. The Laurent series is of the form: $$f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n$$ If $a_n = 0$ for all $n < 0$, then $z_0$ is a removable singularity.

Examples of Removable Singularities

1. Let $f(z) = \frac{\sin z}{z}$. The point $z_0 = 0$ is a singularity because the function is not defined at $z=0$. We can find the limit: $$\lim_{z \to 0} \frac{\sin z}{z} = 1$$ Since the limit exists and is finite, $z_0 = 0$ is a removable singularity. We can define a new function $g(z)$ such that $g(z) = \frac{\sin z}{z}$ for $z \neq 0$ and $g(0) = 1$. This function $g(z)$ is analytic at $z=0$. Alternatively, using the Taylor series for $\sin z$ around $z=0$: $$\sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots$$ So, $$\frac{\sin z}{z} = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots$$ This is a power series (Taylor series) converging for all $z$. The coefficient of $(z-0)^{-1}$ (the principal part) is zero, confirming $z=0$ is removable.
2. Let $f(z) = z \cos\left(\frac{1}{z}\right)$. The point $z_0 = 0$ is a singularity. Consider the Taylor series for $\cos w = 1 - \frac{w^2}{2!} + \frac{w^4}{4!} - \cdots$. Let $w = \frac{1}{z}$. $$\cos\left(\frac{1}{z}\right) = 1 - \frac{1}{2! z^2} + \frac{1}{4! z^4} - \cdots$$ Then, $$f(z) = z \left(1 - \frac{1}{2! z^2} + \frac{1}{4! z^4} - \cdots\right) = z - \frac{1}{2! z} + \frac{1}{4! z^3} - \cdots$$ This Laurent series has terms with negative powers of $z$ (specifically, $\frac{1}{z}$ and $\frac{1}{z^3}$). The limit $\lim_{z \to 0} z \cos\left(\frac{1}{z}\right)$ does not exist (it oscillates between -1 and 1). The function is not bounded near $z=0$. Therefore, $z=0$ is an essential singularity, not removable.
3. Let $f(z) = \frac{e^z - 1}{z}$. The point $z_0 = 0$ is a singularity. The Taylor series for $e^z$ around $z=0$ is $e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots$. So, $e^z - 1 = z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots$. $$\frac{e^z - 1}{z} = 1 + \frac{z}{2!} + \frac{z^2}{3!} + \cdots$$ This is a power series for $z \neq 0$. The limit as $z \to 0$ is 1. Thus, $z=0$ is a removable singularity.

Taylor's Theorem

Taylor's Theorem is a cornerstone of calculus and complex analysis, providing a way to represent functions as infinite sums of terms calculated from the function's derivatives at a single point. In complex analysis, it guarantees that analytic functions can be represented by their Taylor series, which is a powerful tool for analysis and computation.

Taylor's Theorem for Real Functions (Brief Recap)

For a real function $f(x)$ that is $n$ times differentiable at a point $a$, the Taylor polynomial of degree $n$ centered at $a$ is:

$$P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n$$

Taylor's Theorem states that $f(x)$ can be expressed as $P_n(x)$ plus a remainder term $R_n(x)$. If $f(x)$ has derivatives of all orders, and the remainder term approaches 0 as $n \to \infty$, then $f(x)$ can be represented by its Taylor series:

$$f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!} (x-a)^k$$

Taylor's Theorem for Complex Analytic Functions

The complex version of Taylor's Theorem is even stronger because it guarantees the existence of such a series representation for any analytic function within its domain of analyticity.

Statement: If $f(z)$ is analytic in an open domain $D$, and $z_0$ is any point in $D$, then $f(z)$ has a Taylor series expansion around $z_0$ that converges to $f(z)$ in some open disk centered at $z_0$ and contained within $D$. The series is given by:

$$f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n$$ where the coefficients $a_n$ are given by:

$$a_n = \frac{f^{(n)}(z_0)}{n!}$$

So, the Taylor series is:

$$f(z) = f(z_0) + f'(z_0)(z - z_0) + \frac{f''(z_0)}{2!}(z - z_0)^2 + \frac{f'''(z_0)}{3!}(z - z_0)^3 + \cdots$$

This series converges for all $z$ in an open disk $|z - z_0| < R$, where $R$ is the distance from $z_0$ to the nearest singularity of $f(z)$.

Proof Sketch (Using Cauchy's Integral Formula)

Let $f(z)$ be analytic in a domain $D$, and let $z_0 \in D$. Choose a simple closed contour $\gamma$ within $D$ that encloses $z_0$. Let $z$ be any point inside $\gamma$. By Cauchy's Integral Formula for derivatives:

$$f^{(n)}(z_0) = \frac{n!}{2\pi i} \int_{\gamma} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}} d\zeta$$

Now, consider a point $z$ inside the disk bounded by $\gamma$. For any point $\zeta$ on $\gamma$, we have $|\zeta - z_0| > |z - z_0|$. We can write:

$$\frac{1}{\zeta - z} = \frac{1}{(\zeta - z_0) - (z - z_0)} = \frac{1}{\zeta - z_0} \frac{1}{1 - \frac{z - z_0}{\zeta - z_0}}$$

Since $\left|\frac{z - z_0}{\zeta - z_0}\right| < 1$, we can use the geometric series formula $\frac{1}{1-w} = \sum_{n=0}^{\infty} w^n$:

$$\frac{1}{\zeta - z} = \frac{1}{\zeta - z_0} \sum_{n=0}^{\infty} \left(\frac{z - z_0}{\zeta - z_0}\right)^n = \sum_{n=0}^{\infty} \frac{(z - z_0)^n}{(\zeta - z_0)^{n+1}}$$

Now, substitute this into Cauchy's Integral Formula for $f(z)$:

$$f(z) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(\zeta)}{\zeta - z} d\zeta = \frac{1}{2\pi i} \int_{\gamma} f(\zeta) \left( \sum_{n=0}^{\infty} \frac{(z - z_0)^n}{(\zeta - z_0)^{n+1}} \right) d\zeta$$

We can interchange the integral and the summation (this requires justification, but is valid for analytic functions):

$$f(z) = \sum_{n=0}^{\infty} \frac{(z - z_0)^n}{2\pi i} \int_{\gamma} \frac{f(\zeta)}{(\zeta - z_0)^{n+1}} d\zeta$$

Recognizing the integral term as $f^{(n)}(z_0)/n!$:

$$f(z) = \sum_{n=0}^{\infty} \frac{(z - z_0)^n}{2\pi i} \left( \frac{2\pi i f^{(n)}(z_0)}{n!} \right) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!} (z - z_0)^n$$

This completes the proof sketch, showing that the Taylor series converges to $f(z)$ inside the disk.

Examples of Taylor Series

1. $f(z) = e^z$ around $z_0 = 0$ (Maclaurin Series): $f^{(n)}(z) = e^z$ for all $n$. $f^{(n)}(0) = e^0 = 1$. $$e^z = \sum_{n=0}^{\infty} \frac{1}{n!} z^n = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots$$ This series converges for all $z$ (radius of convergence is $\infty$).
2. $f(z) = \frac{1}{1-z}$ around $z_0 = 0$: $f(z)$ is analytic everywhere except at $z=1$. The radius of convergence will be $R=1$. This is already a geometric series: $$\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n = 1 + z + z^2 + z^3 + \cdots$$ This converges for $|z| < 1$.
3. $f(z) = \sin z$ around $z_0 = 0$: $f(0)=0$, $f'(z) = \cos z \implies f'(0)=1$, $f''(z) = -\sin z \implies f''(0)=0$, $f'''(z) = -\cos z \implies f'''(0)=-1$, $f^{(4)}(z) = \sin z \implies f^{(4)}(0)=0$. The derivatives cycle with a period of 4. $$\sin z = 0 + 1(z) + \frac{0}{2!}z^2 + \frac{-1}{3!}z^3 + \frac{0}{4!}z^4 + \frac{1}{5!}z^5 + \cdots$$ $$\sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots$$ This converges for all $z$.
4. $f(z) = \log z$ around $z_0 = 1$: The function $\log z$ is analytic in the domain where $z \neq 0$ and the branch cut is not crossed. Let's use the principal branch. $z_0 = 1$ is in the domain. The nearest singularity is at $z=0$, so $R=1$. $f(1) = \log 1 = 0$. $f'(z) = 1/z \implies f'(1) = 1$. $f''(z) = -1/z^2 \implies f''(1) = -1$. $f'''(z) = 2/z^3 \implies f'''(1) = 2$. $f^{(n)}(z) = (-1)^{n-1} (n-1)! / z^n \implies f^{(n)}(1) = (-1)^{n-1} (n-1)!$ for $n \ge 1$. $$\log z = f(1) + f'(1)(z-1) + \frac{f''(1)}{2!}(z-1)^2 + \frac{f'''(1)}{3!}(z-1)^3 + \cdots$$ $$\log z = 0 + 1(z-1) + \frac{-1}{2!}(z-1)^2 + \frac{2}{3!}(z-1)^3 + \cdots$$ $$\log z = (z-1) - \frac{(z-1)^2}{2} + \frac{(z-1)^3}{3} - \frac{(z-1)^4}{4} + \cdots$$ This series converges for $|z-1| < 1$.