The Argument Principle: Counting Zeros by Winding

Let $f$ be a meromorphic function on a domain containing the closed disk $\overline{D}$ , with no zeros or poles on the boundary circle $\gamma = \partial D$ . Define the logarithmic derivative integral:

$N - P = \frac{1}{2\pi i} \oint_\gamma \frac{f'(z)}{f(z)}\, dz,$

where $N$ is the number of zeros of $f$ inside $D$ (counted with multiplicity) and $P$ is the number of poles (counted with multiplicity).

Problem: Use this principle to determine how many zeros the function

$f(z) = z^4 - 5z + 1$

has inside the annulus $1 < |z| < 2$ .

(Hint: Apply the argument principle on each boundary circle separately, using Rouché's theorem to count zeros on $|z|=1$ and $|z|=2$ separately.)

Answer: Argument Principle: Zeros in an Annulus

Key Idea / Intuition

The argument principle says that $\frac{1}{2\pi i}\oint \frac{f'}{f}\,dz$ counts how many times $f(z)$ winds around the origin as $z$ traverses $\gamma$ — each zero contributes $+1$ winding and each pole contributes $-1$ . To count zeros in an annulus, simply apply Rouché on the outer circle $|z|=2$ and inner circle $|z|=1$ separately, then subtract. Rouché's theorem tells us: if one term dominates on the boundary, the total zero count equals the zero count of the dominant term.

Formal Proof / Solution

Step 1: Zeros inside $|z| < 2$

On $|z| = 2$ , compare $f(z) = z^4 - 5z + 1$ by isolating the dominant term $z^4$ :

$|{-5z + 1}| \leq 5|z| + 1 = 5(2) + 1 = 11,$ $|z^4| = 16.$

Since $16 > 11$ , by Rouché's theorem, $f(z)$ has the same number of zeros inside $|z| < 2$ as $z^4$ , which is $\mathbf{4}$ zeros (with multiplicity).

Step 2: Zeros inside $|z| < 1$

On $|z| = 1$ , compare $f(z) = z^4 - 5z + 1$ by isolating $-5z$ as the dominant term:

$|z^4 + 1| \leq |z|^4 + 1 = 1 + 1 = 2,$ $|-5z| = 5.$

Since $5 > 2$ , by Rouché's theorem, $f(z)$ has the same number of zeros inside $|z| < 1$ as $-5z$ , which is $\mathbf{1}$ zero (at the origin).

Step 3: Zeros in the annulus

The number of zeros of $f$ in the open annulus $1 < |z| < 2$ is:

$(\text{zeros inside } |z|<2) - (\text{zeros inside } |z| \leq 1).$

We need to check: does $f$ have a zero on $|z|=1$ ? If $|z|=1$ and $f(z)=0$ , then $z^4 + 1 = 5z$ , so $|5z| = 5$ but $|z^4+1| \leq 2$ , a contradiction. So no zeros lie on $|z|=1$ .

Therefore:

$\text{zeros in annulus} = 4 - 1 = \boxed{3}.$

Summary of the Argument Principle in action

The logarithmic derivative $\frac{f'}{f}$ has simple poles exactly at the zeros and poles of $f$ , with residue $+\text{ord}$ at zeros and $-\text{ord}$ at poles. Integrating picks up exactly these residues via the residue theorem:

$\frac{1}{2\pi i}\oint_\gamma \frac{f'(z)}{f(z)}\,dz = \sum_{\text{zeros}} \text{ord}(z_k) - \sum_{\text{poles}} \text{ord}(p_k) = N - P.$

This geometric interpretation — counting signed winding number of the image curve $f(\gamma)$ around $0$ — is the heart of the argument principle.