The Argument Principle

For every problem there is always, at least, a solution which seems quite plausible. It is simple and clean, direct, neat and nice, and yet very wrong, Anawim, justtothepoint.com

Recall

Laurent’s theorem. Let A = {$z \in \mathbb{C}: R \lt |z -a| \lt S$} be an annular region ($0 \le R \lt S \le \infty$) and let f be analytic on A. Then, $f(z) = \sum_{n=-\infty}^\infty c_n(z -a)^n, \forall z \in A$ where $c_n = \frac{1}{2\pi i}\int_{\gamma} \frac{f(w)}{(w-a)^{n+1}}dw$ and $\gamma$ is a positively oriented circle of radius r centered at a and R < r < S ($\gamma$ is inside the annular region).

Corollary (aka Rosetta Stone of Complex Analysis). Let f be analytic in $\mathbb{B'}(a; r)$. (i) f has a removable singularity at a if and only if in the Laurent's series expansion of f around a ($\sum_{n = -\infty}^\infty (z-a)^n$) all negative coefficients are zero $c_n = 0, \forall n \lt 0$. (ii) f has a pole of order m ($m \ge 1$) at a if and only if the series stop at the negative power -m, $c_{-m} \ne 0, c_n = 0, \forall n \lt -m$; (iii) f has an isolated essential singularity at a if and only if the series has infinitely many negative terms (it never stops), i.e., there is no m such that $c_n = 0, \forall n \lt -m$

The Argument Principle

The Argument Principle. Suppose f is meromorphic on and inside a simple closed curve $\gamma$ with a finite number of zeroes $a_j, 1 \le j \le l_1$ and poles $b_k, 1 \le k \le l_2$. Suppose none of those zeros and poles lie on $\gamma$. Then, $\frac{1}{2\pi i}\int_{\gamma}\frac{f'(z)}{f(z)}dz = M - N$ where M is the sum of order of zeroes at $a_j, 1 \le j \le l_1$ and N is the sum of order of poles at $b_k, 1 \le k \le l_2$

Proof.

The core of this theorem is the specific function we are integrating: $\frac{f'(z)}{f(z)}$. This is known as the logarithmic derivative because, by the chain rule: $\frac{d}{dz} \ln(f(z)) = \frac{f'(z)}{f(z)}$

1. Logarithms turn powers into multipliers: $\ln(x^n) = n \ln(x)$.
2. Zeros and Poles are defined by powers (e.g., $(z-a_k)^{h_k}$ or $(z-b_j)^{-m_j}$).
3. Therefore, this function converts the order of the zeroes and poles (the power) into a Residue (a coefficient), which the Residue Theorem can “catch.”

Suppose f has a zero of order $h_k$ at $z = a_k$, then we can write the function as f(z) = $(z-a_k)^{h_k}f_1(z)$ where $f_1(z)$ is analytic and $f_1(a_k) \ne 0$.

Now, let’s calculate the fraction $\frac{f'(z)}{f(z)}$. First, we must use the well-known Product Rule to calculate the derivative: $f'(z) = \underbrace{h_k(z-a_k)^{h_k-1}f_1(z)}_{\text{Derivative of first term}} + \underbrace{(z-a)^{h_k} f_1'(z)}_{\text{Derivative of second term}}$

Next, divide this result by the original $f(z)$: $\frac{f'(z)}{f(z)} = \frac{h_k(z-a_k)^{h_k-1}f_1(z)}{(z-a_k)^{h_k} f_1(z)} + \frac{(z-a_k)^{h_k} f_1'(z)}{(z-a_k)^{h_k} f_1(z)}$

Simplify the fractions: $\frac{f'(z)}{f(z)} = \frac{h_k}{(z-a_k)} + \frac{f_1'(z)}{f_1(z)}$. Besides, $\frac{f_1'(z)}{f_1(z)}$ is analytic (no singularity) at $a_k$ and $\frac{h_k}{(z-a_k)}$ is the only singular part of $\frac{f'}{f}$ and represents a simple pole with $Res\{ \frac{f'}{f}, a_k \} = h_k$ (Residue at a Zero = Order of the Zero).

Suppose $f$ has a pole of order $m_j$ at $z = b_j$. We can write the function as: $f(z) = (z-b_j)^{-m_j} f_2(z)$ where $f_2(z)$ is analytic and non-zero at $b_j$.

Using a similar methodology, we could use the Product Rule to calculate $f'(z) = -m_j(z-b_j)^{-m_j-1}f_2(z) + (z-b_j)^{-m_j} f_2'(z)$.

Divide this result by the initial $f(z)$: $\frac{f'(z)}{f(z)} = \frac{-m_j(z-b_j)^{-m_j-1}f_2(z)}{(z-b_j)^{-m_j} f_2(z)} + \frac{(z-b_j)^{-m_j} f_2'(z)}{(z-b_j)^{-m_j} f_2(z)}$

Simplify: $\frac{f'(z)}{f(z)} = \frac{-m_j}{(z-b_j)} + \frac{f_2'(z)}{f_2(z)}$, and $\frac{f_2'(z)}{f_2(z)}$ is analytic (no singularity) at $b_j$ and $\frac{-m_j}{(z-b_j)}$ is the only singular part of $\frac{f'}{f}$ and represents a simple pole with residue $-m_j$, $Res\{ \frac{f'}{f}, b_j \} = -m_j$ (Residue at a Pole = Negative of the Order).

Cauchy’s Residue Theorem can now be applied to the integral of $\frac{f'}{f}$ over the closed curve $\gamma$. The theorem states the integral is $2\pi i \times (\text{Sum of Residues})$, so $\int_{\gamma} \frac{f'(z)}{f(z)}dz = 2\pi i\sum_{k=1}^{l_1}Res\{ \frac{f'}{f}, a_k \} + 2\pi i\sum_{j=1}^{l_2}Res\{ \frac{f'}{f}, b_j \} = 2\pi i(M - N)$.

Each zero $a_k$ contributes its order $h_k$. Total = M. Each pole $b_j$ contributes minus its order $-m_j$. Total = $-N$.

Dividing by $2\pi i$ gives the final result: $\frac{1}{2\pi i} \oint_\gamma \frac{f'(z)}{f(z)} dz = M - N$

Why is this theorem called the Argument Principle? It comes from the fact that the integral counts the total change in the argument (angle) of the function $f(z)$ as $z$ travels around the curve.

Let’s look at the function values $f(z)$ in the image plane (the $w$-plane). Any complex number can be written in polar coordinates: $w = f(z) = Re^{i\theta}$ when R = |f(z)| (the magnitude (radius), a real, positive number representing the distance from the origin) $\theta = arg(f(z))$ (the argument (angle), the angle the point makes with the positive real axis).

Using the product rule and chain rule on $df = e^{i\theta} dR + R(i e^{i\theta}) d\theta = e^{i\theta}(dR + iR d\theta)$

Now, divide by $f(z) = R e^{i\theta}: \frac{df}{f} = \frac{e^{i\theta}(dR + iR d\theta)}{R e^{i\theta}} = \frac{dR}{R} + i d\theta$

Substituting this into the integral over the closed curve $\gamma$: $\frac{1}{2\pi i} \oint_\gamma \frac{f'(z)}{f(z)} dz = \underbrace{\frac{1}{2\pi i} \oint_\gamma \frac{dR}{R}}_{\text{Part A (Magnitude)}} + \underbrace{\frac{1}{2\pi i} \oint_\gamma i d\theta}_{\text{Part B (Angle)}}$

$\text{Part A} = \frac{1}{2\pi i} \oint_\gamma d(\ln R) = \frac{1}{2\pi i} \left[ \ln(R_{final}) - \ln(R_{initial}) \right] = 0$

The magnitude must return to its starting value (change is 0). Mathematically, $|f(z)|$ is a single-valued real function. At any specific point $z$, there is only one specific distance to the origin. Since $z_{start} = z_{end}$ ($\gamma$ is a closed curve), then $R_{start} = R_{end}$.

Consider yourself on a mountain hike. Your altitude is represented by R (magnitude). You begin your journey at a designated location on the trail (e.g., altitude = 500m). You move in a closed loop ($\gamma$). Eventually, your final altitude must be 500 meters, even when you sometimes walk in plain terrain, move up or down at random, and so on.

The only thing that can change is the angle (argument). The integral actually calculates: $\frac{1}{2\pi}\int_{\gamma} d\theta$

$\frac{1}{2\pi}\int_{\gamma} d\theta = n(f(\gamma; \theta)) = \frac{1}{2\pi} (2\pi k) = k.$ This integer k is called the Winding Number of the curve $f(\gamma)$ around the origin. It counts exactly how many times the image wraps around zero.

Imagine you walk around a large tree in the center of a park. $\theta$ represents the direction you are facing or your progress around a central pole. Unlike altitude, angles accumulate! You start facing North ($0^\circ$). You walk a full circle around the tree. When you return to the start point, you are facing North again. However, you have turned $360^\circ$ ($2\pi$ radians). You didn’t just go $0 \to 10 \to 0$. You went $0 \to 180 \to 360$. If you track the continuous change in angle ($\Delta \theta$), it must be a multiple of $2\pi$ (a full rotation): $\Delta \theta_{total} = 2\pi k$ where k is an integer (the number of times you walked around the tree). When you walk around the tree and track the continuous change in your facing angle $\theta$, you’re really measuring how many times your path “winds” around the center.

The theorem says: The value of the integral is determined entirely by the change in the Argument of $f(z)$ along the curve. The number of times the image curve $f(\gamma)$ winds around the origin equals the number of zeros minus the number of poles inside $\gamma$.

$\frac{1}{2\pi} \Delta \theta_{total} =$ # of Zeros - # of Poles = M - N.

• Every Zero inside the curve wraps the image around the origin once in the positive direction ($+2\pi$).
• Every Pole inside the curve wraps the image around the origin once in the negative direction ($-2\pi$).