The Argument Principle and Rouché's Theorem — Examples

On $\mathrm{\mid }z\mathrm{\mid }=1$, $\mathrm{\mid }{z}^5+1\mathrm{\mid }\le \mathrm{\mid }z{\mathrm{\mid }}^5+1=2<3=\mathrm{\mid }3z\mathrm{\mid }\mathrm{.}$

Apply [L1] with $f=p$ and $g(z)=3z$. The function $3z$ has exactly one zero in $\mathrm{\mid }z\mathrm{\mid }<1$, so $p$ has exactly one zero there as well.

On the circle $\mathrm{\mid }z\mathrm{\mid }=2$, $\mathrm{\mid }3z+1\mathrm{\mid }\le 3\mathrm{\mid }z\mathrm{\mid }+1=7<32=\mathrm{\mid }{z}^5\mathrm{\mid }\mathrm{.}$ So [L1] applied to $p$ and $z^5$ shows that $p$ has five zeros in $\mathrm{\mid }z\mathrm{\mid }<2$, counted with multiplicity.

On the circle $\mathrm{\mid }z\mathrm{\mid }=1$, $\mathrm{\mid }{z}^5+1\mathrm{\mid }\le 2<3=\mathrm{\mid }3z\mathrm{\mid }\mathrm{.}$ Another application of [L1] shows that $p$ has exactly one zero in $\mathrm{\mid }z\mathrm{\mid }<1$.

Therefore the number of zeros in the annulus $1<\mathrm{\mid }z\mathrm{\mid }<2$ is $5-1=4$.

If $\mathrm{\mid }z\mathrm{\mid }=1$, then $\mathrm{Re}z\le 1$, so $\mathrm{\mid }{e}^z\mathrm{\mid }=e^{\mathrm{Re}z}\le e<3=\mathrm{\mid }3z\mathrm{\mid }\mathrm{.}$

Apply [L1] to $f(z)=e^z-3z$ and $g(z)=-3z$. Since $-3z$ has exactly one zero in $\mathrm{\mid }z\mathrm{\mid }<1$, so does $f$.

The three zeros of $f$ are the cube roots of unity, so they all lie in $\mathrm{\mid }z\mathrm{\mid }<2$. The function has no poles.

Apply [L1] to the circle $\gamma$. Since $\gamma$ is null-homologous in $\mathbb{C}$ and encloses all three zeros of $f$, the argument-principle count is $3$. Therefore $n(f\circ \gamma ,0)=3$.

On every compact set, $\mathrm{\mid }{z}^2\mathrm{/}(n+1)\mathrm{\mid }\to 0$, so $f_n\to f$ locally uniformly. The limit function $f(z)=z$ has a simple zero at $0$.

Apply [L1] to the isolated zero at $0$. It follows that on some disc $D(0,r)$, every sufficiently large $f_n$ has exactly one zero counted with multiplicity.

On the circle $\mathrm{\mid }z\mathrm{\mid }=1\mathrm{/}4$ one has $\mathrm{\mid }{z}^2-w\mathrm{\mid }\le \mathrm{\mid }z{\mathrm{\mid }}^2+\mathrm{\mid }w\mathrm{\mid }<\frac{1}{16}+\frac{1}{8}=\frac{3}{16}<\frac{1}{4}=\mathrm{\mid }z\mathrm{\mid }\mathrm{.}$ So Rouché's theorem applied to $z$ and $z+z^2-w$ shows that $f(z)-w$ has exactly one zero inside $\gamma$.

The two roots of $z^2+z-w$ are $(-1\pm s)\mathrm{/}2$. Since $\mathrm{Re}s>0$, one has $\mathrm{\mid }s+1\mathrm{\mid }\ge 1+\mathrm{Re}s>1$, so $\mathrm{\mid }s-1\mathrm{\mid }=\frac{\mathrm{\mid }{s}^2-1\mathrm{\mid }}{\mathrm{\mid }s+1\mathrm{\mid }}=\frac{4\mathrm{\mid }w\mathrm{\mid }}{\mathrm{\mid }s+1\mathrm{\mid }}<\frac{1}{2}\mathrm{.}$ Therefore $\mid \frac{-1+s}{2}\mid <\frac{1}{4},\mid \frac{-1-s}{2}\mid =\frac{\mathrm{\mid }1+s\mathrm{\mid }}{2}>\frac{1}{2}>\frac{1}{4}\mathrm{.}$ So the unique zero inside $\gamma$ is $z(w)=(-1+s)\mathrm{/}2$.

If $f(z)=w$ and $\mathrm{\mid }z\mathrm{\mid }<1\mathrm{/}4$, then $\mathrm{\mid }1+2z\mathrm{\mid }\ge 1-2\mathrm{\mid }z\mathrm{\mid }>1\mathrm{/}2$, so the enclosed zero is simple. The positively oriented circle $\gamma (t)=\frac{1}{4}{e}^{it}$ winds once around every point of $\mathrm{\mid }z\mathrm{\mid }<1\mathrm{/}4$, so the hypotheses of [L1] are satisfied and the inverse contour formula gives $z(w)=\frac{1}{2\pi i}{\int }_{\gamma }\frac{\zeta (1+2\zeta )}{\zeta +{\zeta }^2-w}d\zeta \mathrm{.}$

Every $f_n(z)=z\mathrm{/}(n+1)$ is entire and injective. On each compact set, $\mathrm{\mid }{f}_n(z)\mathrm{\mid }\le {\sup }_K\mathrm{\mid }z\mathrm{\mid }\mathrm{/}(n+1)\to 0$, so $f_n\to 0$ locally uniformly.

The limit function is the constant $0$, which is not injective. Thus the claim without the constant escape clause is false, exactly as [L1] warns.

The equation $e^{1\mathrm{/}z}-1=0$ is equivalent to $1\mathrm{/}z=2\pi ik$ for some nonzero integer $k$, so the zeros are $z_k=\frac{1}{2\pi ik}(k\in \mathbb{Z}\setminus \{0\})\mathrm{.}$ These are infinitely many distinct points, and they accumulate at $0$.

Thus every small circle around $0$ encloses infinitely many zeros of $e^{1\mathrm{/}z}-1$, while $0$ is an essential singularity rather than a pole. The finite meromorphic count required by [L1] is unavailable, so the naive extension of the argument principle to essential singularities is false.

On $\mathrm{\mid }z\mathrm{\mid }=1$ one has $\mathrm{\mid }f(z)-g(z)\mathrm{\mid }=\mathrm{\mid }1\mathrm{\mid }=1=\mathrm{\mid }z\mathrm{\mid }=\mathrm{\mid }g(z)\mathrm{\mid },$ so the weak inequality holds everywhere on the boundary.

But $f$ vanishes at $z=-1$, which lies on the boundary itself. So the interior zero-count conclusion is no longer even well posed. This is exactly why [L1] is stated with a strict inequality.

Each $f_n(z)=z\mathrm{/}(n+1)$ is injective, entire, and converges locally uniformly to the constant function $0$.

The limit $0$ is not injective. Therefore the claim is false, and [L1] identifies the missing clause exactly.

The function $f(z)=z^2$ has exactly one distinct zero, namely $0$, but that zero has multiplicity $2$.

Applying [L1] on the unit circle gives $\frac{1}{2\pi i}{\int }_{\gamma }\frac{f^{\mathrm{\prime }}(z)}{f(z)}dz=\frac{1}{2\pi i}{\int }_{\gamma }\frac{2}{z}dz=2.$ So the argument principle returns $2$, not the number of distinct zeros. The claim is false.