6.7 Counting Zeroes

Zeroes of An Analytic Function
A zero of a function \(f\) is a point \(a\) where \(f(a) = 0\).

Define \(z(t) = \big \{z\in G:\, f(z) = 0\big \}\)
Suppose \(f\) is analytic on \(B(a:r),\,r>0\) and suppose \(f(a)= 0\). Taylor’s theorem tells us that \(\displaystyle {f(z) = \sum ^{\infty }_{n=0} C_n \big (z - a\big )^n}\) for \(\left |z - a\right |< r\) where \(C_n = \frac {1}{n!}f^n(a)\,,\, C_0 = 0\) since \(f(a) = 0\)

Case 1:: If all the derivatives of \(f\) at \(a\) are zero then \(f\) is identically the zero function on \(B(a;r)\).
Case 2:: If not, let \(m\) be the smallest positive integer such that \(f^m(a) \neq 0\). Then
\(\displaystyle {f(z) = \big (z - a\big )^m\sum ^{\infty }_{n = m} C_n \big (z - a\big )^{n - m}}\) for \(z \in B(a;r)\) with the first term \(C_m \neq 0\), in this case \(f\) is said to have a zero of order \(m\) at \(a\). The zero is called a simple zero if \(m= 1\).

Example 6.32. \(\sin (z) = 0\) for \(z = k\pi \implies \) \(\frac {d}{dz}\sin z = \cos z \neq 0\) at \(z =k\pi \). So all the zeros of \(\sin z\) are simple

\(\implies \quad z\sin z\) has a zero of order 2 at \(z = 0\).

If \(f\) and \(g\) are analytic on \(B(a,r)\) and \(f(a) = g(a) = 0\) and if \(hf\) is of the order of zero of \(f\) at \(a\) and \(hg\) is the order zero of \(f\cdot g\) at \(a\) has order \(hf + hg\).

From above, let \(\displaystyle {f(z) = (z - a)^m\sum ^{\infty }_{n=m}C_n(z - a)^{n-m}}\) where \(C_n \neq 0\) for \(z\in B(a;r)\). Define \(\displaystyle {\Psi (z) = \sum ^{\infty }_{n=m}C_n (z - a)^{n-m}}\) on \(B(a;r),\quad \Psi (a) \neq 0\). Since \(\Psi \) is a converging power series on \(B(a;r),\, \Psi \) is analytic and hence continuous. So there is an \(\varepsilon > 0\) such that \(\Psi (z) \neq 0\) for \(z\in B(a;\varepsilon )\).
So we conclude that the zeroes of \(f\) are isolated in this case.

Theorem 6.33 (Identity Theorem). Suppose that \(f\) is analytic on \(B(a;r),\, r> 0\) and that \(f(a) = 0\). Then either

1.: \(f\) is identically zero in \(B(a;r)\) or
2.: then zero of \(f\) at \(a\) is isolated i.e \(\exists \, \varepsilon > 0 \) such that the punctured disc \(B'(a;\varepsilon )\) has no other zero of \(f\)

Proof. Expand \(f\) in its Taylor series about \(a\), valid on the disc: \[f(z)=\sum _{n=0}^{\infty }a_n(z-a)^n,\qquad a_n=\frac {f^{(n)}(a)}{n!} .\] Since \(f(a)=0\) we have \(a_0=0\). Two cases arise.

If every \(a_n=0\), the series is identically zero and so is \(f\) on \(B(a;r)\) — alternative (1).

Otherwise let \(m\geq 1\) be the least index with \(a_m\neq 0\). Factoring, \[f(z)=(z-a)^m g(z),\qquad g(z)=\sum _{k=0}^{\infty }a_{m+k}(z-a)^{k},\] where \(g\) is analytic on the disc and \(g(a)=a_m\neq 0\). By continuity there is \(\varepsilon >0\) with \(g(z)\neq 0\) for \(\left |z-a\right |<\varepsilon \). On the punctured disc \(0<\left |z-a\right |<\varepsilon \) both factors are non-zero, so \(f(z)\neq 0\) — alternative (2), the zero at \(a\) is isolated.

The dichotomy is sharp: an analytic function cannot have zeros piling up at a point of its domain without vanishing identically. This is what makes the next theorem possible. □

Consequently if \(a\) is a limit point of \(\zeta (f)\) then \(f\equiv 0\) in \(B(a;r).\, \)
\(\Big [\) Every \(B\Big (a;\frac {1}{n}\Big )\) contains a zero of \(f\). So \((2)\) cannot happen which implies \((1)\) has to happen\(\Big ]\).

Identity theorem: (General theorem)
Let \(G\) be a region and suppose that \(f\) is analytic on \(G\). Assume that the set \(\zeta (f)\) of the zeros of \(f\) has a limit point in \(G\). Then \(f\) is identically zero in \(G\).

Theorem 6.34 (Uniqueness Theorem). Let \(G\) be a region and suppose that \(f\) and \(g\) are analytic functions on \(G\). Suppose that \(f(z) = g(z),\,\forall z \in S\), where \(S\) has a limit point in \(G\), then \(f(z) \equiv g(z)\) on \(G\).

Now suppose \(f\) is analytic on \(B(a;r)\) with \(f(a) = 0\). Suppose \(f\neq 0\) on \(B(a;r)\) and \(f(z)\neq 0\) on \(B'(a;r)\). Then \(f(z) = (z - a)^m\Psi (z)\) for \(z \in B(a;r)\) where \(\Psi (a)\neq 0\). So \[f'(z) = m(z - a)^{m-1}\Psi (z) + (z- a)^m\Psi '(z)\,\text {for}\, z \in B(a;r)\] Since \(\Psi (z) \neq 0\) in \(B(a;\varepsilon )\) for some \(\varepsilon > 0\), for \(z \neq a, \, z\in B(a;\varepsilon )\), we have \[\frac {f'(z)}{f(z)} = \frac {m}{z - a} + \frac {\Psi ' (z)}{\Psi (z)}\] If \(C{\varepsilon _0}\) is a contour whose trace is a circle of radius \(\varepsilon _0\) ( where \(\varepsilon > \varepsilon _0 > 0\)) with centre \(a\), oriented in the positive direction, then \[\int _{C_{\varepsilon _0}}\frac {f'(z)}{f(z)}\,dz = \int _{C_{\varepsilon _0}} \frac {m}{z - a}\,dz +\underbrace {\int _{C_{\varepsilon _0}}\frac {\Psi '(z)}{\Psi (z)}\,dz}_0\]

\[\implies \quad \frac {1}{2\pi i}\int _{C_{\varepsilon _0}}\frac {f'(z)}{f(z)}\,dz = m\]

Proof. Put \(h=f-g\), analytic on \(G\), and let \(S\) be the set on which \(h\) vanishes. By hypothesis \(S\) has a limit point \(a\in G\).

Then \(h(a)=0\) by continuity, since \(a\) is a limit of points where \(h\) vanishes. But the zero of \(h\) at \(a\) is not isolated — every punctured disc about \(a\) contains points of \(S\). By the previous theorem the only remaining possibility is that \(h\equiv 0\) on some disc \(B(a;\rho )\).

To spread this over all of \(G\), let \[A=\{z\in G: h\equiv 0 \text { on a neighbourhood of } z\} .\] \(A\) is open by construction, and non-empty since \(a\in A\). It is also closed in \(G\): if \(z_k\in A\) with \(z_k\rightarrow z\in G\), then \(z\) is a non-isolated zero of \(h\), so by the previous theorem \(h\) vanishes on a disc about \(z\) and \(z\in A\). A non-empty subset of the connected set \(G\) that is both open and closed must be all of \(G\). Hence \(h\equiv 0\), that is \(f\equiv g\), on \(G\).

Connectedness is doing real work here: on a disconnected open set the two functions could agree on one component and differ on another. □

Theorem 6.35 (Counting Zeroes). Let \(f\) be analytic inside and on a positively oriented contour \(\Gamma \). Let \(f\) be non-zero on \(\Gamma ^*\) and have \(N\) zeroes inside \(\Gamma (\) including multiplicity of zeroes \()\). Then \[\frac {1}{2\pi i}\int _{\Gamma } \frac {f'(z)}{f(z)}\,dz = N\]

Proof. The function \(\frac {f'}{f}\) is analytic inside and on \(\Gamma \) except at the zeroes of \(f\) lying inside \(\Gamma \). Suppose the zeroes are \(a_1, a_2, \ldots , a_n\) of orders \(m_1, m_2, \ldots , m_n\) inside \(\Gamma \). We can find disjoint open discs \(B(a_k,r_k)\, k = 1, \cdots n\) such that there is a function \(\Psi _k\) which is analytic and non-zero \(B(a_k,r_k)\) and such that \(f(z) = (z - a_k)^{m_k}\Psi _k(z)\) for \(z \in B(a_k, r_k)\). Then \[\frac {f'(z)}{f(z)} = \frac {m_k}{z - a_k} + \frac {\Psi _k'(z)}{\Psi _k(z)}\quad (z\in B'(a_k,r_k))\]

Define \(\,\displaystyle {F(z) = \begin {cases} \frac {f'(z)}{f(z)} - \sum \limits ^n_{j=1}\frac {m_j}{z - a_j} & \text {if}\quad z\in \bigcup ^n_{k = 1}B(a_k;r_k)\\\\ \frac {\Psi _k'(z)}{\Psi _k(z)} - \sum \limits _{\substack {1\leq j\leq n\\ j\neq k}}\frac {m_j}{z - a_j} & \text {if}\quad z \in B(a_k;r_k)\, 1\leq k\leq n \end {cases} }\\\quad F\) is analytic.

Now \(\displaystyle {\int _{\Gamma } F(z)dz = 0}\) by Cauchy’s theorem.

i.e \(\quad \displaystyle {\int _{\Gamma } \frac {f'(z)}{f(z)} \,dz - \int _{\Gamma } \sum ^n_{j= 1} \frac {m_j}{z - a_j}\,dz = 0}\)

\(\implies \quad \displaystyle {\frac {1}{2\pi i}\int _{\Gamma } \frac {f'(z)}{f(z)}\,dz = N\quad \Bigg (\therefore \quad \sum ^n_{j= 1}m_j = N}\Bigg )\) □

Theorem 6.36 (Rouche’s Theorem). Let \(f(z)\) and \(g(z)\) be analytic inside and on a contour \(\Gamma \) and suppose that \(\left |f(z)\right |> \left |g(z)\right |\) on \(\Gamma ^*\). Then \(f(z)\) and \(f(z) + g(z)\) have the same number of zeroes inside \(\Gamma \).

Proof. Let \(t\in \big [0, 1\big ]\). Since \(\left |f(z)\right |>\left |g(z)\right |\) on \(\Gamma ^*\, \big (f + tg\big ) (z) \neq 0\) for \(z \in \Gamma ^*\). Orient \(\Gamma \) positively \[\Psi (t) = \frac {1}{2\pi i}\int _{\Gamma } \frac {\big (f' + tg'\big )(z)}{\big (f + tg\big )}\,dz\] \(\Psi (t)\) is the number of zeroes of \(f + tg\) inside \(\Gamma \). So \(\Psi \) is integral valued. If \(\Psi \) is continuous then \(\Psi \) must be a constant function.
\(\Psi (0)\) is the number of zeroes of \(f\) inside \(\Gamma \) and \(\Psi (1)\) is the number of zeroes of \(f + g\) inside \(\Gamma \).
\(\Psi \) is continuous: Fix \(t\) \[\Psi (t) - \Psi (s) = \frac {t - s}{2\pi i}\int _{\Gamma }\frac {\big (g'f - f'g\big ) (z)}{\big ( f + tg\big )(z)\big (f + sg\big )(z)}\,dz\] There are positive constants \(M\) and \(m\) such that \(\forall z \in \Gamma ^*\), \(\,\left |g' f - f' g\right |\leq M,\quad \left |g(z)\right |\leq M\) and \(\left |\big (f + tg\big )\right |\geq m\). Then \begin {align*} \left |\big (f + sg\big )(z)\right | & = \left |\big (f + tg\big )(z) + \big (s - t\big )g(z)\right |\\ & \geq \left |\big (f + tg\big )(z)\right | - \left |s - t\right | \left |g(z)\right |\\ & \geq \frac {1}{2}m \quad \text {if}\quad \left |s - t\right |\leq \frac {m}{2M} \end {align*}

Hence for \(\left |s - t\right |\) small enough. \[\left |\Psi (t) - \Psi (s)\right | \leq \frac {\left |t - s\right |\,M}{\pi \, m^2}\times \text {length}\big (\Gamma \big )\] So \(\Psi \) is continuous. □

Example 6.37. \(f(z) = 2 + z^2 - e^{iz}\,, \quad \Big ( e^{iz} = z^2 + 2\Big )\)

We will show that there is precisely one zero of \(f\) in the upper half plane.

Let \(f_1(z) = 2 + z^2 \) and let \(g_1(z) = -e^{iz}\). Then \(\big (f_1 + g_1\big )(z) = f(z)\) on \(\big [-\mathbb {R},\mathbb {R}\big ]\)

\begin {align*} \left |f_1(z)\right | & = \left |2 + z^2\right | \geq 2 > 1 = \left |g_1(z)\right |\\\\ \implies \quad \left |f_1(z)\right | & > \left |g_1(z)\right | \end {align*}

Also on \(Re^{i\theta }\, (0\leq \theta \leq \pi )\) \[\left |f_1(z)\right |\leq \underbrace {R^2 - 2}_{R>\sqrt {3}}> 1 \geq e^{-R\sin \theta } = \left |g_1(z)\right |\]

\begin {align*} g_1(z) & = -e^{i\big (R\cos \theta + iR\sin \theta \big )}\\ & = -e^{-R\sin \theta + iR\cos \theta } \end {align*}

\(\left |f_1(z)\right | > \left |g_1(z)\right |\) on \( Re^{i\theta }\)
By Rouche’s theorem the number of zeroes of \(f_1(z)\) is equal to the number of zeroes of \(f(z)\) inside \(\Gamma \), \(f_1(z)\) has only one zero in the upper half plane, \(f(z)\) also has only one zero in the upper half plane.

Example 6.38. How many zeroes of the polynomial \(g(z) = 2z^5 - 5z^2 + 43\) lie inside the circle \(\left |z\right | = 2?\)

Example 6.39. Show that \(g(z) = z^5 - 4z^4 + 4z^2 - 7\) has exactly \(4\) zeros inside \(\left |z\right | = 2\).

Now \(f(z)=z^4(z-4)\) has a zero of order \(4\) at the origin and a simple zero at \(z=4\), outside. So \(f\) has \(4\) zeros inside \(\left |z\right |=2\), and therefore so does \(g\).

Computing the roots of \(g\) numerically confirms it: four have moduli \(1.079\), \(1.079\), \(1.266\), \(1.266\), and the fifth sits at \(z\approx 3.751\).

Corollary 6.40 (Open Mapping Theorem). A non-constant analytic function maps open sets to open sets.
(Let \(f\) be a non-constant analytic function on an open set \(G\). Then \(f(G)\) is open in \(\mathbb {C})\).

Proof. Let \(f\) be a non-constant analytic function on an open set \(G\). Let \(a\in G\), the \(f(a)\in f(G)\).
We must show that \(f(G)\) is open. There is a \(\delta > 0 \ni B\big (f(a),\delta \big ) \subset f(G)\) and every
\(w\in B\big (f(a),\delta \big )\) has a pre-image in \(G\) via \(f\).
Choosing sufficiently small \(\varepsilon \) there is a \(\delta > 0 \ni \) every value of \(B\big (f(a),\delta \big )\) is assumed the same number of times inside a circle of radius \(\varepsilon \) around \(a\).
So \(B\big (f(a);\delta \big )\subseteq f\big (B(a;\varepsilon )\big ) \subset f(G)\). So \(f\) is open map. □

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