Complex Analysis

2 Entire functions

The number of zeroes of an entire function is related to the growth of the function.

Theorem 2.1 (Jensen’s Formula). Let \(f\) be holomorphic near the disk of radius \(R\), and nonzero at \(0\). Then if \(z_1,\dots , z_k\) are the zeros of \(f\), \(\log |f(0)| = \sum \log (\frac {|z_i|}{R}) + \frac {1}{2 \pi } \int _0^{2\pi } \log |f(Re^{i\theta })|d\theta \).

Proof. Note that if the formula holds for \(f,g\), then it holds for \(fg\). Thus we only need to show it for \(z-z_0\) where \(z_0\) lies in the disk, and non-vanishing holomorphic functions. For the latter, note that \(\log |f(z)|\) is the real part of the log of \(f\), so the formula follows from the mean value property. For \(f=z-z_0\), we have \(\frac {1}{2 \pi } \int \log |Re^{i\theta }-z_0|d\theta - \log R = \frac {1}{2 \pi } \int \log |e^{i\theta }-\frac {z_0}{R}|d\theta = \frac {1}{2 \pi } \int \log |1-\frac {z_0}{R}e^{i \theta }|d\theta \) by a change of variables \(\theta \to -\theta \), which by the mean value property is \(\real (\log (1-\frac {z_0}{R}z))|_{z = 0} = 0\). □

Given a holomorphic function \(f\) not vanishing at \(0\), let \(n_f(r)\) be the number of zeros with modulus \(\leq R\). Then Jensen’s formula gives

Corollary 2.2. \(\int _0^R\frac {n_f(r)}{r}dr = \frac {1}{2 \pi }\int _0^{2\pi } \log |f(Re^{i \theta })|d\theta - \log |f(0)|\).

As an example, let \(f\) be an entire function of finite order \(\rho \), where \(\rho \) is the infimum of \(r\) such that \(|f(z)| \leq Ce^{A|z|^r}\).

Theorem 2.3. If \(f\) is of finite order \(\leq \rho \), then \(n_f(R) = O(|R|^\rho )\). Moreover, if \(z_k\) are the zeros, then \(\sum _{i}|z_k|^{-s}\) converges for \(s > \rho \).

Proof. WLOG, \(f(0) \neq 0\). Jensen’s formula gives \(n_f(R) \leq 2\int _R^{2R}\frac {n_f(r)}{r}dr \leq \frac {1}{\pi }\int \log |f(2Re^{i\theta })|d\theta - \log |f(0)| = O(|R|^\rho )\). This bound generally implies convergence of the series. □

Lemma 2.4. Let \(F_i(z)\) be holomorphic functions on \(U\) such that \(|F_i(z)-1|\leq c_n\) for some summable sequence \(c_n\). Then \(F(z) = \prod _i F_i(z)\) converges uniformly to a holomorphic function on \(U\) that vanishes iff one of the \(F_i\) does. Moreover, if no \(F_i\) vanishes at a point \(p\), then \(\frac {F'(p)}{F(p)} = \sum _i \frac {F_i'(p)}{F_i(p)}\).

Proof. WLOG we can assume that \(|F_i(z)-1| \leq \frac {1}{2}\) as \(c_n\) is summable. Then we have \(F(z) = e^{\sum _i\log F_i(z)}\). By the power series for \(\log (1+z)\), \(\log F_i(z) \leq 2|F_i(z)-1| \leq 2c_n\), so the sum uniformly converges to a holomorphic function. Note in particular this shows that \(F(z)\) doesn’t vanish as it is the exponential of something finite. Moreover, \(\log (F)'(z) = \sum _i\log (F_i)'(z) = \sum _i \frac {F'}{F}(z)\). □

We will now derive the product formula for \(\sin (z)\).

Lemma 2.5. \(\pi \cot (\pi z) = \sum _{k \in \ZZ }\frac {1}{z+k} = \frac {1}{z}+\sum _{n>0}\frac {2z}{z^2-n^2}\).

Proof. These can be characterized by Liouville’s theorem as the unique \(\ZZ \)-periodic meromorphic function bounded away from a simple pole at \(\ZZ \) of residue \(1\). □

Proposition 2.6. \(\sin (\pi z) = \pi z \prod _1^\infty (1-\frac {z^2}{n^2})\).

Proof. Their logarithmic derivatives agree by the previous two lemmas, and so the two sides agree up to multiplication by a constant. By dividing by \(z\) and taking the limit as \(z \to 1\), we get \(1\) on both sides, so they are equal. □

Given a sequence \(a_n \in \CC \) approaching \(\infty \), we can produce a function \(F\) vanishing exactly at the \(a_i\). WLOG, the \(a_i\) are nonzero. The function \(\prod (1-\frac {z}{a_i})\) may not converge, so we will have to correct for this. The idea is as follows: \((1-z) = e^{\log (1-z)} = e^{-\sum \frac {z^j}{j}}\). Thus \(E_k(z) = (1-z)e^{{\sum _1^k}\frac {z^j}{j}}\). for large \(k\) will not grow as fast, but will have the same zeros as \(E_0(z) = (1-z)\). The \(E_k(z)\) are called canonical factors.

Theorem 2.7 (Weierstrass). If \(a_i \neq 0\) tend to \(\infty \), \(\prod _iE_i(\frac {z}{a_i})\) is entire and has zeroes exactly at the \(a_i\).

Proof. For \(|z| \leq \frac {1}{2}\), \(|E_i(z)-1| = |e^{\sum _{k+1}^\infty \frac {z^j}{j}}-1|\). The sum is bounded above by \(c|z|^{k+1}\), so we get this is \(\leq c'|z^{k+1}|\). If we fix some \(r\), then in the ball of radius \(r\), eventually the \(\frac {z}{a_i}\) are within \(\frac {1}{2}\), so the product converges. □

Any function \(f\) can be divided by a function constructed as above to get a non-vanishing entire function, which must be of the form \(e^{g(z)}\) for some entire \(g\). This can be refined for \(f\) of finite order \(\rho \).

Lemma 2.8. Let \(g\) be a holomorphic function such that \(\real g(z) \leq c|z|^s\) for values of \(|z|\) tending to \(\infty \). Then \(g\) is a polynomial of degree at most \(s\).

Proof. Let \(g(z) = \sum _i a_i z^i\). By Lemma 1.6, we have that \(a_n= \frac {1}{2 \pi R^n}\int _0^{2\pi } {g(Re^{i\theta })}{e^{-in\theta }}d\theta \). Conjugating the equation and adding for \(n\) and \(-n\) gives \(a_n= \frac {1}{\pi R^n}\int _0^{2\pi } {\real g(Re^{i\theta })}{e^{-in\theta }}d\theta \) for \(n>0\) and \(\real a_0 = \frac {1}{\pi }\int _0^{2\pi } {\real g(e^{i\theta })}d\theta \). Thus we have for \(n>s\), \(|a_n|= \frac {1}{\pi R^n}|\int _0^{2\pi } {\real g(Re^{i\theta })}{e^{-in\theta }}d\theta | = \frac {1}{\pi R^n}|\int _0^{2\pi } {\real (cR^s-g(Re^{i\theta }))}{e^{-in\theta }}d\theta | \) \(\leq 2CR^{s-n}-\real a_0 R^{-n} \to 0\) as \(R \to \infty \). □

Theorem 2.9 (Hadamard factorization theorem). If \(f\) is a holomorphic function of finite order \(\rho \) and \(k = \floor \rho \), then \(f = e^P(z) z^m \prod _i E_k(\frac {z}{a_i})\), where \(P\) is a polynomial of degree \(\leq k\).

Proof. The product \(E(z) = z^m \prod _i E_k(\frac {z}{a_i})\) converges, where \(m\) is the order of the zero of \(f\), and \(a_i\) the other zeroes, as \(f\) is finite order, so \(|E_k(\frac z{a_i})-1|\) is bounded from above by \(c|\frac {z}{a_i}|^{k+1}\), and the series \(\frac {1}{|a_i|^{k+1}}\) converges. Thus \(\frac {f}{E}\) is an entire function with no zeroes, so is \(e^{g(z)}\) for some \(g\). If we can produce lower bounds on \(E\) for arbitrarily large \(|z|\), then by the lemma, \(g\) will have to be a polynomial. To do this, for \(\rho <s<k+1\), we would like to show that \(|\prod _k E_k(\frac {z}{a_n})| \geq e^{-C|z|^s}\) except maybe in the balls around each \(a_i\) of radius \(|a_i|^{-(k+1)}\). Then since \(R-\sum _{|a_i| \leq R} 2|a_i|^{-k+1} \to \infty \) as \(R \to \infty \), there are are arbitrarily large \(R\) such that when \(|z|=R\), it doesn’t intersect any of the balls, and the bound holds. This bound will then be good enough to show that \(g\) is a polynomial of the right degree.

To obtain the bound, first obtain from the definition of \(E_k(z)\) that when \(|z| \leq \frac {1}{2}\), \(|E_k(z)| \geq e^{-c|z|^{k+1}}\) and \(|E_k(z)| \geq |1-z|e^{-c'|z|^{k}}\) if \(|z| \geq \frac {1}{2}\). Then split the product \(|\prod _k E_k(\frac {z}{a_n})|\) into the product when \(|a_n| \leq 2|z|\) and \(|a_n|>2|z|\). For the second product, we have \(\prod _n |E_k(\frac {z}{a_n})| \geq e^{-c|\sum _n\frac {|z|^{k+1}}{a_n^{k+1}}}\) \(\geq e^{-c''|z|^s|\sum _n{a_n^{-s}}} \geq e^{-c'''|z|^s|} \) as the sum converges. For the first product, \(\prod _n |E_k(\frac {z}{a_n})| \geq \prod |1-\frac {z}{a_n}| \prod e^{-c'|\frac {z}{a_n}|^k}\). The first term is \(\prod \frac {|a_n-z|}{|a_n|}\), which is \(\geq \prod \frac {1}{|a_i|^{k+2}} \geq e^{-(k+2)\log n_f(2|z|)\log (2|z|)}\), and applying Theorem 2.3, this is \(\geq e^{-C'|z|^{s+\ee }}\), where \(\ee >0\) is arbitrary. Note that if we had just chosen \(s\) a bit smaller, then \(s+\ee \) can be chosen to be the original real number we wanted. Finally for \(\prod e^{-c'|\frac {z}{a_n}|^k}\), note \(c'\sum |\frac {z}{a_n}|^k \leq C''|z|^s\sum \frac {1}{|a_n|^s}\), so that the overall bound \(|\prod _k E_k(\frac {z}{a_n})| \geq e^{-C|z|^s}\) holds outside the chosen balls. □