§3.3 Zeros of a Holomorphic Function

3.3 Zeros of a Holomorphic Function

For a region $𝑈 \subseteq ℂ$ and a holomorphic function $𝑓 : 𝑈 \to ℂ$ , a point $𝑧_{0} \in 𝑈$ is a zero of $𝑓$ iff $𝑓 (𝑧_{0}) = 0$ . Furthermore, if $𝑓$ has the Taylor expansion at $𝑧_{0}$ of

𝑎_{𝑚} {(𝑧 - 𝑧_{0})}^{𝑚} + 𝑎_{𝑚 + 1} {(𝑧 - 𝑧_{0})}^{𝑚 + 1} + \dots, 𝑚 \in ℕ, 𝑎_{𝑚} \neq 0,

then the zero at $𝑧_{0}$ has multiplicity $𝑚$ .

We will introduce a fundamental application of Liouville’s Theorem (Theorem 3.2.5 $Theorem 3.2.5 Theorem 3.2.5$ ) below.

Theorem 3.3.1 (Fundamental Theorem of Algebra).

Every non-constant polynomial

𝑝 (𝑧)

with complex coefficients has at least one complex zero.

Proof.

For the sake of contradiction, suppose that

𝑝 (𝑧)

has no complex zeros. Then the function

𝑓 (𝑧) = \frac{1}{𝑝 (𝑧)}

is continuous and entire, because

𝑝 (𝑧)

has no zeros in

ℂ

. Moreover, as

𝑧 \to \infty

𝑝 (𝑧) \to \infty

, so

𝑓 (𝑧) \to 0

, and thus

𝑓 (𝑧)

is bounded. By Liouville’s Theorem (Theorem 3.2.5

Theorem 3.2.5 Theorem 3.2.5

), every bounded entire function is constant. Thus,

𝑓 (𝑧)

is constant, and so

𝑝 (𝑧)

must also be constant. By contradiction,

𝑝 (𝑧)

has at least one complex zero. ⁠

□

Theorem 3.3.2.

Let $𝑈 \subseteq ℂ$ be open and connected, and $𝑓 : 𝑈 \to ℂ$ be holomorphic over $𝑈$ . Then if the set defined by

𝑆 = {𝑧 \in 𝑈 : 𝑓 (𝑧) = 0}

has an accumulation point in

𝑈

, then

𝑓 \equiv 0

over

𝑈

Proof.

Let ${𝑧_{𝑛}}_{𝑛 \in ℕ}$ be a subset of $𝑆$ and assume it has an accumulation point $𝑧_{\infty}$ in $𝑈$ . Since $𝑓$ is holomorphic over $𝑈$ , $\exists 𝜀 > 0$ such that $𝑓$ is holomorphic over $𝐷 (𝑧_{\infty}, 𝜀) \subseteq 𝑈$ . Then over this disk, $𝑓$ has the Taylor expansion

𝑓 (𝑧) = \sum_{𝑛 = 0}^{\infty} 𝑎_{𝑛} {(𝑧 - 𝑧_{\infty})}^{𝑛} . (3.3 . 1)

By Definition 1.1.3 $Definition 1.1.3 Definition 1.1.3$ , $\exists 𝑁 \in ℕ$ such that $\forall 𝑛 > 𝑁$ , $𝑧_{𝑛} \in 𝐷 (𝑧_{\infty}, 𝜀)$ . Since $𝑧_{𝑛}$ is a zero of $𝑓$ , $𝑓 (𝑧_{𝑛}) = 0$ . Then, by the continuity of $𝑓$ ,

lim_{𝑛 \to \infty} 𝑓 (𝑧_{𝑛}) = 𝑓 (lim_{𝑛 \to \infty} 𝑧_{𝑛}) = 𝑓 (𝑧_{\infty}) = 0 .

Using this result in comparison to (3.3.1) $(3.3.1) (3.3.1)$ , we get that $𝑎_{0} = 0$ .

The function

𝑓_{1} (𝑧) = \frac{𝑓 (𝑧)}{𝑧 - 𝑧_{\infty}}

has a Taylor expansion over $𝐷 (𝑧_{\infty}, 𝜀)$ of

𝑓_{1} (𝑧) = \sum_{𝑛 = 0}^{\infty} 𝑎_{𝑛 + 1} {(𝑧 - 𝑧_{\infty})}^{𝑛} .

Let $𝑧 = 𝑧_{𝑛} \neq 𝑧_{\infty}$ for some $𝑛 > 𝑁$ . Then $𝑓_{1}$ vanishes, leaving

0 = 𝑎_{1} + 𝒪︀ (𝑧_{𝑛} - 𝑧_{\infty}) .

Letting $𝑛 \to \infty$ , $𝑧_{𝑛} \to 𝑧_{\infty}$ , and $𝑎_{1} = 0$ . Define

𝑓_{2} (𝑧) = \frac{𝑓_{1} (𝑧)}{𝑧 - 𝑧_{\infty}} .

Then,

𝑓_{2} (𝑧) = \sum_{𝑛 = 0}^{\infty} 𝑎_{𝑛 + 2} {(𝑧 - 𝑧_{\infty})}^{𝑛} .

Similarly, $𝑎_{2} = 0$ . Letting

𝑓_{𝑛} (𝑧) = \frac{𝑓 (𝑧)}{{(𝑧 - 𝑧_{\infty})}^{𝑛}},

the sequence ${𝑎_{𝑛}}_{𝑛 \in ℤ_{\geq 0}}$ vanishes, and $𝑓 \equiv 0$ on $𝐷 (𝑧_{\infty}, 𝜀)$ .

Let

\tilde{𝑆} = {𝑧 \in 𝑈 : \forall 𝑛 \in ℤ_{\geq 0}, 𝑓^{(𝑛)} (𝑧) = 0} .

For all $𝑧 \in 𝐷 (𝑧_{\infty}, 𝜀)$ , since $𝑓 (𝑧)$ locally vanishes (and has vanishing derivatives as a consequence),

𝐷 (𝑧_{\infty}, 𝜀) \subseteq \tilde{𝑆} .

Furthermore, for all $𝑧^{'} \in \tilde{𝑆}$ , $\exists 𝜀^{'} > 0$ such that $𝑓 (𝑧)$ has a convergent Taylor series with vanishing coefficients on $𝐷 (𝑧^{'}, 𝜀^{'}) \subseteq 𝑈$ . Then $𝑓 \equiv 0$ on $𝐷 (𝑧^{'}, 𝜀^{'})$ . Then for all $𝑧 \in 𝐷 (𝑧^{'}, 𝜀^{'})$ , since $𝑓$ is constant at $𝑧$ , it also has vanishing derivatives. It follows that

𝐷 (𝑧^{'}, 𝜀^{'}) \subseteq \tilde{𝑆} .

Since every point in $\tilde{𝑆}$ has an open neighborhood also in $\tilde{𝑆}$ , $\tilde{𝑆}$ is open.

It is evident that for all $𝑘 \in ℤ_{\geq 0}$ , $𝑓^{(𝑘)}$ is continuous in $𝑈$ by the holomorphy of $𝑓$ . Let

𝑆_{𝑘} = {𝑧 \in 𝑈 : 𝑓^{(𝑘)} (𝑧) = 0} .

For any sequence ${{\tilde{𝑧}}_{𝑛}} \in 𝑆_{𝑘}$ converging to some ${\tilde{𝑧}}_{\infty} \in 𝑈$ , by the continuity of $𝑓$ ,

lim_{𝑛 \to \infty} 𝑓^{(𝑘)} ({\tilde{𝑧}}_{𝑛}) = 𝑓^{(𝑘)} (lim_{𝑛 \to \infty} {\tilde{𝑧}}_{𝑛}) = 𝑓^{(𝑘)} ({\tilde{𝑧}}_{\infty}) = 0,

and therefore ${\tilde{𝑧}}_{\infty} \in 𝑆_{𝑘}$ . Thus, $𝑆_{𝑘}$ contains all of its accumulation points in $𝑈$ and is therefore closed in $𝑈$ (if ${\tilde{𝑧}}_{\infty} \notin 𝑈$ , then it is no longer relevant; we are concerned about it being closed within $𝑈$ ). Since

\tilde{𝑆} = ⋂_{𝑘 \in ℤ_{\geq 0}} 𝑆_{𝑘}

and each of $𝑆_{𝑘}$ is closed in $𝑈$ , $\tilde{𝑆}$ is the intersection of closed sets and consequently closed.

Since $\tilde{𝑆}$ is nonempty and clopen in the connected set $𝑈$ , $\tilde{𝑆} = 𝑈$ (by Theorem 3.2.1.19 $Theorem 3.2.1.19 Theorem 3.2.1.19$ ). It follows that $𝑓 \equiv 0$ on $𝑈$ . ⁠ $□$

Remark.

This is a trivial property of holomorphic functions that allows for the uniqueness of analytic continuations. It is oftentimes stated in the form below:

Theorem 3.3.3 (Identity Theorem).

Let

𝑈 \subseteq ℂ

be open and connected, and define

𝑓 (𝑧)

and

𝑔 (𝑧)

to be two holomorphic functions on

𝑈

. For a set

𝑆 \subseteq 𝑈

with an accumulation point in

𝑈

, if

𝑓 \equiv 𝑔

𝑆

, then

𝑓 \equiv 𝑔

𝑈

Proof.

Let

ℎ = 𝑓 - 𝑔

be holomorphic over

𝑈

. Since

𝑆

has an accumulation point in

𝑈

, and

ℎ \equiv 0

over

𝑆

, then by Theorem 3.3.2

Theorem 3.3.2 Theorem 3.3.2

ℎ \equiv 0

over

𝑈

. ⁠

□

Theorem 3.3.4 (Holomorphic Argument Principle).

Let $𝑈 \subseteq ℂ$ be a region and $𝑓 : 𝑈 \to ℂ$ be holomorphic. Let $𝛾 \subset 𝑈$ be a simple, closed, positively oriented curve that is null-homotopic in $𝑈$ . If $𝑓$ has no zeros on $𝛾$ , then $𝑓$ has finitely many zeros in the region bounded by $𝛾$ , and this number, counting multiplicities, is given by

𝑘 = \frac{1}{2 π i} \oint_{Γ} \frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)} d 𝑧 .

Let $Γ$ be the image of $𝛾$ under the map $𝑤 = 𝑓 (𝑧)$ . Then

𝑘 = \frac{1}{2 π} Δ_{Γ} arg (𝑤),

where $Δ_{Γ} arg (𝑤)$ denotes the total change in argument of $𝑤$ as it traverses $Γ$ .

Proof.

Let $𝑧_{1}, \dots, 𝑧_{𝑛}$ be the distinct zeros of $𝑓$ enclosed by $𝛾$ with the respective multiplicities $𝑘_{1}, \dots, 𝑘_{𝑛}$ . Choose disjoint disks $𝐷 (𝑧_{𝑗}, 𝜀_{𝑗})$ centered at each $𝑧_{𝑗}$ with radii $𝜀_{𝑗} > 0$ , each contained in the interior of $𝛾$ and avoiding $𝛾$ . The function

\frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)}

is holomorphic on the domain

int (𝛾) ∖ ⋃_{𝑗 = 1}^{𝑛} \overline{𝐷 (𝑧_{𝑗}, 𝜀_{𝑗})},

where $int (𝛾)$ denotes the interior relative to $𝛾$ . The oriented boundary of this domain is

𝛾^{+} \cup ⋃_{𝑗 = 1}^{𝑛} 𝜕 {𝐷 (𝑧_{𝑗}, 𝜀_{𝑗})}^{-} .

By Cauchy–Goursat (Theorem 3.1.9 $Theorem 3.1.9 Theorem 3.1.9$ ),

\int_{𝛾^{+} \cup ⋃_{𝑗 = 1}^{𝑛} 𝜕 {𝐷 (𝑧_{𝑗}, 𝜀_{𝑗})}^{-}} \frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)} d 𝑧 = 0,

which rearranges to

\oint_{𝛾^{+}} \frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)} d 𝑧 = \sum_{𝑗 = 1}^{𝑛} \oint_{𝜕 {𝐷 (𝑧_{𝑗}, 𝜀_{𝑗})}^{+}} \frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)} d 𝑧 .

Near each $𝑧_{𝑗}$ , express

𝑓 (𝑧) = {(𝑧 - 𝑧_{𝑗})}^{𝑘_{𝑗}} ℎ_{𝑗} (𝑧)

where $ℎ_{𝑗}$ is holomorphic and non-vanishing on $𝐷 (𝑧_{𝑗}, 𝜀_{𝑗})$ . Differentiation yields

𝑓^{'} (𝑧) = 𝑘_{𝑗} {(𝑧 - 𝑧_{𝑗})}^{𝑘_{𝑗} - 1} ℎ_{𝑗} (𝑧) + {(𝑧 - 𝑧_{𝑗})}^{𝑘_{𝑗}} ℎ_{𝑗}^{'} (𝑧),

and thus

\frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)} = \frac{𝑘_{𝑗}}{𝑧 - 𝑧_{𝑗}} + \frac{ℎ_{𝑗}^{'} (𝑧)}{ℎ_{𝑗} (𝑧)} .

Since $ℎ_{𝑗}$ is holomorphic and non-vanishing on $𝐷 (𝑧_{𝑗}, 𝜀_{𝑗})$ , the function $ℎ_{𝑗}^{'} / ℎ_{𝑗}$ is holomorphic there. By the Cauchy–Goursat Theorem,

\oint_{𝜕 𝐷 (𝑧_{𝑗}, 𝜀_{𝑗})} \frac{ℎ_{𝑗}^{'} (𝑧)}{ℎ_{𝑗} (𝑧)} d 𝑧 = 0 .

The Cauchy–Goursat Formula (Theorem 3.1.10 $Theorem 3.1.10 Theorem 3.1.10$ ) gives

\oint_{𝜕 𝐷 (𝑧_{𝑗}, 𝜀_{𝑗})} \frac{𝑘_{𝑗}}{𝑧 - 𝑧_{𝑗}} d 𝑧 = 2 π i 𝑘_{𝑗} .

Combining results,

\oint_{Γ} \frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)} d 𝑧 = \sum_{𝑗 = 1}^{𝑛} 2 π i 𝑘_{𝑗} = 2 π i 𝑘 .

Finally, parameterize $Γ$ by $𝑤 = 𝑓 (𝑧)$ . Then $d 𝑤 = 𝑓^{'} (𝑧) d 𝑧$ , and

𝑘 = \frac{1}{2 π i} \oint_{Γ} \frac{d 𝑤}{𝑤} = \frac{1}{2 π i} Δ_{Γ} log (𝑤) = \frac{1}{2 π} Δ_{Γ} arg (𝑤),

which proves the result. ⁠ $□$

Thus, one defines the winding index ( $Ind$ ) to quantify how many times a closed curve winds counterclockwise around a given point in the complex plane. Formally, if $𝛾 = 𝛾 ([0, 1])$ is a counterclockwise-oriented closed curve and $𝑧$ is a point satisfying $𝑧 \notin 𝛾$ , then

{Ind}_{Γ} (𝑧) = \frac{1}{2 π i} \oint_{𝛾} \frac{d 𝜁}{𝜁 - 𝑧} = \frac{1}{2 π i} \int_{0}^{1} \frac{𝛾^{'} (𝑡) d 𝑡}{𝛾 (𝑡) - 𝑧} .

Theorem 3.3.5.

Let

{𝑓_{𝑛} (𝑧)}

be a sequence of holomorphic functions on the open set

𝑈 \subseteq ℂ

that uniformly converges to

𝑓 (𝑧)

on every compact subset of

𝑈

. If

\forall 𝑛 \in ℕ

𝑓_{𝑛} (𝑧)

has no zeros in

𝑈

, then

𝑓

is either identically

0

or has no zeros in

𝑈

Proof.

By the holomorphy of $𝑓_{𝑛} (𝑧)$ , for any simple closed rectifiable curve $𝛾 \subset 𝑈$ (whose interior is a subset of $𝑈$ ), by the Cauchy–Goursat Theorem (Theorem 3.1.9 $Theorem 3.1.9 Theorem 3.1.9$ ),

\oint_{Γ} 𝑓_{𝑛} (𝜁) d 𝜁 = 0 .

Since $𝛾$ is a subset of any compact subset of $𝑈$ , ${𝑓_{𝑛} (𝜁)}$ uniformly converges on $𝛾$ , and by Theorem 2.3.7 $Theorem 2.3.7 Theorem 2.3.7$ ,

lim_{𝑛 \to \infty} \oint_{Γ} 𝑓_{𝑛} (𝜁) d 𝜁 = \oint_{Γ} lim_{𝑛 \to \infty} 𝑓_{𝑛} (𝜁) d 𝜁 = \oint_{Γ} 𝑓 (𝜁) d 𝜁 = 0 . (3.3 . 2)

Then by Morera’s Theorem (Theorem 3.2.6 $Theorem 3.2.6 Theorem 3.2.6$ ), $𝑓 (𝑧)$ is holomorphic, and $𝑓^{'} (𝑧)$ is holomorphic. We aim to show that $𝑓_{𝑛}^{'} (𝑧) ⇉ 𝑓^{'} (𝑧)$ .

Let $𝐾 ⋐ 𝑈$ be arbitrary and compact and $𝑉 \supset 𝐾$ be open and relatively compact in $𝑈$ . Since ${𝑓_{𝑛}^{'} (𝑧)}$ is holomorphic, by Corollary 3.2.7.1 $Corollary 3.2.7.1 Corollary 3.2.7.1$ , there exists a finite constant $𝑐 > 0$ such that

lim_{𝑛 \to \infty} sup_{𝑧 \in 𝐾} | 𝑓_{𝑛}^{'} (𝑧) - 𝑓^{'} (𝑧) | \leq 𝑐 lim_{𝑛 \to \infty} sup_{𝑧 \in 𝑉} | 𝑓_{𝑛} (𝑧) - 𝑓 (𝑧) | .

By the definition of uniform convergence, the right-hand side approaches $0$ , and ${𝑓_{𝑛}^{'} (𝑧)}$ is then uniformly convergent to $𝑓^{'} (𝑧)$ by the same reasoning.

Through the proof of Theorem 3.3.2 $Theorem 3.3.2 Theorem 3.3.2$ , if $𝑓 ≢ 0$ over $𝑈$ , then the zeros of $𝑓$ do not have an accumulation point in $𝑈$ and are therefore discrete. In this case, let $𝛾 \subset 𝑈$ be a curve that does not pass through the zeros of $𝑓$ . Since each function in the sequence $𝑓_{𝑛}$ does not contain zeros in $𝑈$ , by the Argument Principle (Theorem 3.3.4 $Theorem 3.3.4 Theorem 3.3.4$ ),

lim_{𝑛 \to \infty} \oint_{Γ} \frac{𝑓_{𝑛}^{'} (𝑧)}{𝑓_{𝑛} (𝑧)} d 𝑧 = 0 . (3.3 . 3)

Since $𝑓$ and $𝑓^{'}$ are holomorphic over $𝛾$ , by Theorem 1.2.22 $Theorem 1.2.22 Theorem 1.2.22$ , there exists a finite value $𝑀 > 0$ such that $\forall 𝑧 \in 𝛾$ , $max {| 𝑓 (𝑧) |, | 𝑓^{'} (𝑧) |} < 𝑀$ .

Since $𝛾$ does not pass through the zeros of $𝑓$ , $\exists 𝜆 > 0$ such that $\forall 𝑧 \in 𝛾$ , $| 𝑓 (𝑧) | > 𝜆$ . By the uniform convergence of ${𝑓_{𝑛} (𝑧)}$ , $\exists 𝑁 \in ℕ$ such that

| 𝑓_{𝑛} (𝑧) - 𝑓 (𝑧) | < \frac{𝜆}{2}, \forall 𝑛 > 𝑁, \forall 𝑧 \in 𝛾 .

Then $| 𝑓_{𝑛} (𝑧) | > \frac{𝜆}{2}$ on $𝛾$ . Hence, $\frac{1}{𝑓_{𝑛} (𝑧)}$ and its limit are uniformly bounded;

| \frac{1}{𝑓 (𝑧)} | < \frac{1}{𝜆}, | \frac{1}{𝑓_{𝑛} (𝑧)} | < \frac{2}{𝜆}, \forall 𝑧 \in 𝛾, \forall 𝑛 > 𝑁 .

\begin{aligned} | \frac{𝑓^{'}}{𝑓} - \frac{𝑓_{𝑛}^{'}}{𝑓_{𝑛}} | & = | \frac{𝑓^{'} 𝑓_{𝑛} - 𝑓_{𝑛}^{'} 𝑓}{𝑓_{𝑛} 𝑓} | \\ < 2 \frac{| 𝑓^{'} 𝑓_{𝑛} - 𝑓^{'} 𝑓 | + | 𝑓^{'} 𝑓 - 𝑓_{𝑛}^{'} 𝑓 |}{𝜆^{2}} \\ < \frac{2 𝑀}{𝜆^{2}} \cdot (| 𝑓_{𝑛} - 𝑓 | + | 𝑓^{'} - 𝑓_{𝑛}^{'} |) . \end{aligned}

\begin{aligned} sup_{𝑧 \in 𝛾} | \frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)} - \frac{𝑓_{𝑛}^{'} (𝑧)}{𝑓_{𝑛} (𝑧)} | & \leq \frac{2 𝑀}{𝜆^{2}} (sup_{𝑧 \in 𝛾} | 𝑓_{𝑛} (𝑧) - 𝑓 (𝑧) | + sup_{𝑧 \in 𝛾} | 𝑓^{'} (𝑧) - 𝑓_{𝑛}^{'} (𝑧) |) \\ lim_{𝑛 \to \infty} sup_{𝑧 \in 𝛾} | \frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)} - \frac{𝑓_{𝑛}^{'} (𝑧)}{𝑓_{𝑛} (𝑧)} | & \leq \frac{2 𝑀}{𝜆^{2}} (lim_{𝑛 \to \infty} sup_{𝑧 \in 𝛾} | 𝑓_{𝑛} (𝑧) - 𝑓 (𝑧) | + | 𝑓^{'} (𝑧) - 𝑓_{𝑛}^{'} (𝑧) |) \\ = 0 . \end{aligned}

Therefore, $\frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)}$ is uniformly convergent on $𝛾$ . By Theorem 2.3.7 $Theorem 2.3.7 Theorem 2.3.7$ , we can pass the limit through the integral in (3.3.3) $(3.3.3) (3.3.3)$ . Then,

lim_{𝑛 \to \infty} \oint_{𝛾} \frac{𝑓_{𝑛}^{'} (𝑧)}{𝑓_{𝑛} (𝑧)} d 𝑧 = \oint_{𝛾} \frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)} d 𝑧 = 0 .

By the Argument Principle (Theorem 3.3.4 $Theorem 3.3.4 Theorem 3.3.4$ ), $𝑓 (𝑧)$ has no zeros in the interior of $𝛾$ . Since $𝛾$ was arbitrarily chosen, either $𝑓 (𝑧) \equiv 0$ on $𝑈$ or has no zeros in $𝑈$ . ⁠ $□$

Theorem 3.3.6 (Rouché).

Let $𝑈 \subseteq ℂ$ be open and $𝑓, 𝑔$ be two holomorphic functions over $𝑈$ . Let $𝛾 \subset 𝑈$ be a simple, closed, rectifiable curve, and for all $𝑧 \in 𝛾$

| 𝑓 (𝑧) - 𝑔 (𝑧) | < | 𝑓 (𝑧) | . (3.3 . 4)

Then $𝑓$ and $𝑔$ have the same number of zeros enclosed by $𝛾$ and do not vanish on $𝛾$ .

Proof.

It is obvious that $𝑔 (𝑧)$ has no zeros on $𝛾$ . Otherwise, $\exists 𝑧_{0} \in 𝛾$ such that $𝑔 (𝑧_{0}) = 0$ , implying that $| 𝑓 (𝑧_{0}) | < | 𝑓 (𝑧_{0}) |$ which is impossible. Similarly, $𝑓 (𝑧)$ has no zeros on $𝛾$ , since $| 𝑔 (𝑧) | < 0$ is an impossibility.

Let $𝑘_{𝑓}$ and $𝑘_{𝑔}$ denote the number of zeros of $𝑓$ and $𝑔$ enclosed by $𝛾$ , respectively. By the Argument Principle (Theorem 3.3.4 $Theorem 3.3.4 Theorem 3.3.4$ ),

\begin{aligned} 𝑘_{𝑔} - 𝑘_{𝑓} & = \oint_{Γ} \frac{𝑔^{'} (𝑧)}{𝑔 (𝑧)} d 𝑧 - \oint_{𝛾} \frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)} d 𝑧 \\ = \oint_{𝛾} \frac{𝑔^{'} (𝑧) 𝑓 (𝑧) - 𝑓^{'} (𝑧) 𝑔 (𝑧)}{𝑔 (𝑧) 𝑓 (𝑧)} d 𝑧 = \oint_{𝛾} \frac{{(\frac{𝑔 (𝑧)}{𝑓 (𝑧)})}^{'}}{\frac{𝑔 (𝑧)}{𝑓 (𝑧)}} d 𝑧 . \end{aligned}

Let $𝑤 = ℎ (𝑧) = \frac{𝑔 (𝑧)}{𝑓 (𝑧)}$ with $Γ = ℎ (𝛾)$ . Then,

𝑘_{𝑔} - 𝑘_{𝑓} = \oint_{Γ} \frac{d 𝑤}{𝑤} .

From (3.3.4) $(3.3.4) (3.3.4)$ , by dividing both sides by $𝑓 (𝑧)$ , we obtain $| 𝑤 - 1 | < 1$ . Then $Γ$ lies in the open disk $𝐷 (1, 1)$ , which will never intersect or enclose $0$ . Then by Lemma 3.1.8 $Lemma 3.1.8 Lemma 3.1.8$ ,

𝑘_{𝑔} - 𝑘_{𝑓} = \oint_{Γ} \frac{d 𝑤}{𝑤} = 0,

as desired. ⁠ $□$

By the Fundamental Theorem of Algebra (Theorem 3.3.1 $Theorem 3.3.1 Theorem 3.3.1$ ), any polynomial in the form $𝑝 (𝑧) = \sum_{𝑘 = 0}^{𝑛} 𝑎_{𝑘} 𝑧^{𝑘}$ ( $𝑛 \in ℕ$ , $𝑎_{𝑛} \neq 0$ , $𝑎_{𝑘} \in ℂ$ where $𝑘 = 1, \dots 𝑛$ ) has at least one complex zero. Consider the function $𝑞 (𝑧) = 𝑎_{𝑛} 𝑧^{𝑛}$ , with a zero at $𝑧 = 0$ with multiplicity $𝑛$ . By Rouché’s Theorem (Theorem 3.3.6 $Theorem 3.3.6 Theorem 3.3.6$ ), since $\exists 𝑅 \in ℝ$ such that $| 𝑞 (𝑧) - 𝑝 (𝑧) | = | \sum_{𝑘 = 0}^{𝑛 - 1} 𝑎_{𝑘} 𝑧^{𝑘} | < | 𝑎_{𝑛} 𝑧^{𝑛} |$ over $| 𝑧 | = 𝑅$ , $𝑝$ and $𝑞$ have the same number of zeros, counting multiplicity.

Theorem 3.3.7.

Let $𝑈 \subseteq ℂ$ be open and connected, and $𝑓 (𝑧)$ be holomorphic and non-constant on $𝑈$ .

If $𝑧_{0} \in 𝑈$ and $𝑤_{0} = 𝑓 (𝑧_{0})$ , and the multiplicity of the zero at $𝑧_{0}$ of $𝑓 - 𝑤_{0}$ is $𝑚$ , then for all $𝜌 > 0$ such that $𝑓 - 𝑤_{0}$ is non-vanishing on $\overline{𝐷 (𝑧_{0}, 𝜌)} ∖ {𝑧_{0}}$ , $\exists 𝛿 > 0$ such that $\forall 𝜉 \in 𝐷 (𝑤_{0}, 𝛿)$ , $𝑓 - 𝜉$ has $𝑚$ zeros in $𝐷 (𝑧_{0}, 𝜌)$ , counting multiplicity.

Proof.

The zero at $𝑧_{0}$ is isolated by Theorem 3.3.2 $Theorem 3.3.2 Theorem 3.3.2$ . Furthermore, $| 𝑓 - 𝑤_{0} |$ is continuous on $𝜕 𝐷 (𝑧_{0}, 𝜌)$ and attains a positive infimum $𝛿$ . In other words, on this set, $| 𝑓 - 𝑤_{0} | \geq 𝛿$ . Hence, $\forall 𝜉 \in 𝐷 (𝑤_{0}, 𝛿)$ , we have $| 𝜉 - 𝑤_{0} | < 𝛿 \leq | 𝑓 (𝑧) - 𝑤_{0} |$ for any $𝑧 \in 𝜕 𝐷 (𝑧_{0}, 𝜌)$ .

By Rouché’s Theorem, since $| (𝑓 (𝑧) - 𝑤_{0}) - (𝑓 (𝑧) - 𝜉) | < | 𝑓 (𝑧) - 𝑤_{0} |$ , it follows that $𝑓 - 𝜉$ and $𝑓 - 𝑤_{0}$ have the same number of zeros in $𝐷 (𝑧_{0}, 𝜌)$ . ⁠ $□$

We also have the following generalization of Theorem 3.3.5 $Theorem 3.3.5 Theorem 3.3.5$ , which is a heuristic restatement of Theorem 3.3.7 $Theorem 3.3.7 Theorem 3.3.7$ :

Theorem 3.3.8 (Hurwitz).

Let $𝑈 \subseteq ℂ$ be an open and connected set, and suppose ${𝑓_{𝑛} (𝑧)}_{𝑛 \in ℕ}$ is a holomorphic function sequence that uniformly converges to a non-constant function $𝑓 (𝑧)$ on all compact sets of $𝑈$ .

If $𝑧_{0} \in 𝑈$ and $𝑤_{0} = 𝑓 (𝑧_{0})$ , and the multiplicity of the zero at $𝑧_{0}$ of $𝑓 - 𝑤_{0}$ is $𝑚$ , then for all $𝜌 > 0$ such that $𝑓 - 𝑤_{0}$ is non-vanishing on $\overline{𝐷 (𝑧_{0}, 𝜌)} ∖ {𝑧_{0}}$ , $\exists 𝑁 \in ℕ$ such that $\forall 𝑛 > 𝑁$ , $𝑓_{𝑛} - 𝑤_{0}$ has $𝑚$ zeros in $𝐷 (𝑧_{0}, 𝜌)$ , counting multiplicity.

Proof.

The zero at $𝑧_{0}$ is isolated by Theorem 3.3.2 $Theorem 3.3.2 Theorem 3.3.2$ . Furthermore, $| 𝑓 - 𝑤_{0} |$ is continuous on $𝜕 𝐷 (𝑧_{0}, 𝜌)$ and attains a positive infimum $𝛿$ . In other words, on this set, $| 𝑓 - 𝑤_{0} | \geq 𝛿$ . By uniform convergence, $\exists 𝑁 \in ℕ$ such that $\forall 𝑛 > 𝑁$ , we have $| 𝑓 (𝑧) - 𝑓_{𝑛} (𝑧) | < 𝛿 \leq | 𝑓 (𝑧) - 𝑤_{0} |$ for any $𝑧 \in 𝜕 𝐷 (𝑧_{0}, 𝜌)$ .

By Rouché’s Theorem (Theorem 3.3.6 $Theorem 3.3.6 Theorem 3.3.6$ ), since

| (𝑓 (𝑧) - 𝑤_{0}) - (𝑓_{𝑛} (𝑧) - 𝑤_{0}) | < | 𝑓 (𝑧) - 𝑤_{0} |,

it follows that $𝑓_{𝑛} - 𝑤_{0}$ and $𝑓 - 𝑤_{0}$ have the same number of zeros in $𝐷 (𝑧_{0}, 𝜌)$ . ⁠ $□$