06.01.17 · riemann-surfaces / complex-analysis

Weierstrass factorization theorem

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Anchor (Master): Weierstrass 1876 *Zur Theorie der eindeutigen analytischen Functionen*; Hadamard 1893 *Etude sur les proprietes des fonctions entieres*; Ahlfors *Complex Analysis* Ch. 5; Conway *Functions of One Complex Variable* Ch. VII; Boas *Entire Functions*

Intuition [Beginner]

A polynomial can be factored into linear pieces: $z^{2} - 1 = (z - 1) (z + 1)$ , and $z^{3} - z = z (z - 1) (z + 1)$ . Each factor corresponds to a zero of the polynomial, and the factorisation records where the zeros are and what power they appear with.

The Weierstrass factorization theorem says that every entire function (a function that is smooth in the complex sense everywhere on the plane) can be factored the same way. If the function has zeros at points $a_{1}, a_{2}, a_{3}, \dots$ (possibly infinitely many), then the function equals a product of factors, one for each zero, multiplied by a never-zero function of the form $e^{g (z)}$ .

The complication is convergence. A product of infinitely many factors might not converge. Weierstrass solved this by introducing correction factors: instead of the raw factor $(1 - z / a_{n})$ for each zero $a_{n}$ , he used "primary factors" that converge in a controlled way. The correction ensures the infinite product always converges to an entire function with exactly the prescribed zeros.

Why does this concept exist? The fundamental theorem of algebra says every polynomial factors by its zeros. Weierstrass extended this from polynomials to all entire functions, providing the single most important structural theorem in the theory of entire functions.

Visual [Beginner]

A diagram showing the zeros of the function $sin (π z)$ on the complex plane: dots at each integer $z = 0, 1, - 1, 2, - 2, \dots$ along the real axis. The Weierstrass product for $sin (π z)$ is displayed as $π z$ times the product of factors $(1 - z / n)$ and $(1 + z / n)$ for $n = 1, 2, 3, \dots$ (after grouping $n$ and $- n$ ).

The picture shows that the zeros of $sin (π z)$ are exactly the integers, and the Weierstrass product reconstructs the entire function from these zeros alone.

Worked example [Beginner]

The function $sin (π z)$ has zeros at every integer: $z = 0, 1, - 1, 2, - 2, \dots$ . Construct the Weierstrass product step by step.

Step 1. The zero at $z = 0$ contributes the factor $z$ . The zero at $z = 1$ contributes $(1 - z /1) = (1 - z)$ . The zero at $z = - 1$ contributes $(1 - z / (- 1)) = (1 + z)$ .

Step 2. Grouping the pair $n$ and $- n$ for each positive integer $n$ : $(1 - z / n) (1 + z / n) = 1 - z^{2} / n^{2}$ . The product through $N$ pairs is $z (1 - z^{2}) (1 - z^{2} /4) (1 - z^{2} /9) \dots (1 - z^{2} / N^{2})$ .

Step 3. As $N \to \infty$ , this product converges to $sin (π z) / π$ . Multiplying by $π$ : $sin (π z) = π z \cdot (1 - z^{2} / 1^{2}) (1 - z^{2} / 2^{2}) (1 - z^{2} / 3^{2}) \dots$ . This is the Euler product for $sin (π z)$ , a special case of the Weierstrass factorisation.

What this tells us: the sine function is completely determined by its zeros (the integers) and the normalising constant $π$ . The Weierstrass machinery produces the function from its zeros.

Check your understanding [Beginner]

Formal definition [Intermediate+]

The Weierstrass primary factors are defined for $p = 0, 1, 2, \dots$ by

E_{0} (z) = 1 - z, E_{p} (z) = (1 - z) exp (z + \frac{z ^{2}}{2} + \dots + \frac{z ^{p}}{p}) (p \geq 1) .

Each $E_{p} (z)$ is an entire function with a single simple zero at $z = 1$ and no other zeros. The exponential correction factor $exp (z + z^{2} /2 + \dots + z^{p} / p)$ is designed to make $∣ E_{p} (z) - 1∣$ small when $∣ z ∣$ is small: the Taylor expansion of $lo g (1 - z)$ begins $- z - z^{2} /2 - \dots$ , and the first $p$ terms of this expansion are cancelled by the exponential factor.

Definition (Canonical product). Given a sequence ${a_{n}}$ of non-zero complex numbers with $∣ a_{n} ∣ \to \infty$ and a non-negative integer $m$ (the multiplicity of the zero at the origin), the canonical product is

P (z) = z^{m} n = 1 \prod \infty E_{p} (\frac{z}{a _{n}}),

where $p$ is chosen so that $\sum_{n = 1}^{\infty} (∣ z ∣/∣ a_{n} ∣)^{p + 1}$ converges. ^{[Ahlfors Ch. 5]}

Counterexamples to common slips

The naive product $(1 - z / a_{1}) (1 - z / a_{2}) \dots$ does not always converge. The product $\prod (1 - z / a_{n})$ converges if and only if $\sum ∣ z / a_{n} ∣$ converges. For sequences like $a_{n} = n$ (the zeros of $sin (π z)$ ), this sum diverges, so the raw product diverges. The Weierstrass primary factors correct this.
The genus $p$ depends on the density of zeros. If the zeros grow like $∣ a_{n} ∣ \sim n^{1/ ρ}$ , the genus is $p = ⌊ ρ ⌋$ . Faster zero accumulation requires higher genus.
The exponential factor $e^{g (z)}$ is essential. The product captures the zeros but not the growth rate or phase. The factor $e^{g (z)}$ accounts for everything about the function that is not determined by its zeros.

Key theorem with proof [Intermediate+]

Theorem (Weierstrass factorization). Let $f$ be an entire function with a zero of order $m$ at the origin and zeros at $a_{1}, a_{2}, a_{3}, \dots$ (counted with multiplicity, $0 < ∣ a_{1} ∣ \leq ∣ a_{2} ∣ \leq \dots$ and $∣ a_{n} ∣ \to \infty$ ). Then there exists an entire function $g$ such that

f (z) = e^{g (z)} \cdot z^{m} \cdot n = 1 \prod \infty E_{p} (\frac{z}{a _{n}}),

where $p$ is chosen so that $\sum_{n} ∣ a_{n} ∣^{- (p + 1)} < \infty$ and the product converges uniformly on compact subsets of $C$ .

Proof. Choose $p$ so that $\sum ∣ a_{n} ∣^{- (p + 1)} < \infty$ (such $p$ exists because $∣ a_{n} ∣ \to \infty$ ). Define the canonical product

P (z) = z^{m} n = 1 \prod \infty E_{p} (\frac{z}{a _{n}}) .

To show convergence: the key estimate is $∣ E_{p} (w) - 1∣ \leq ∣ w ∣^{p + 1}$ for $∣ w ∣ \leq 1$ . This follows from

lo g E_{p} (w) = lo g (1 - w) + w + \frac{w ^{2}}{2} + \dots + \frac{w ^{p}}{p} = - k = p + 1 \sum \infty \frac{w ^{k}}{k},

so $∣ lo g E_{p} (w) ∣ \leq ∣ w ∣^{p + 1} / (p + 1) \cdot 1/ (1 - ∣ w ∣) \leq ∣ w ∣^{p + 1}$ for $∣ w ∣ \leq 1/2$ . A sharper bound gives $∣ E_{p} (w) - 1∣ \leq c ∣ w ∣^{p + 1}$ for $∣ w ∣ \leq 1$ . Since $\sum ∣ z / a_{n} ∣^{p + 1} \leq ∣ z ∣^{p + 1} \sum ∣ a_{n} ∣^{- (p + 1)} < \infty$ for each fixed $z$ (and the tail of the sum is controlled by the convergent series), the product converges uniformly on compact sets.

Both $f (z)$ and $P (z)$ are entire functions with the same zeros at the same multiplicities. Hence $f (z) / P (z)$ is an entire function with no zeros. An entire function with no zeros can be written as $e^{g (z)}$ for some entire function $g$ : since $f / P$ is entire and non-vanishing, $lo g (f / P)$ is well-defined (locally, by the existence of a holomorphic logarithm on simply-connected domains; globally, by the monodromy theorem since $C$ is simply connected). Set $g (z) = lo g (f (z) / P (z))$ . Then $f (z) = e^{g (z)} P (z)$ . $□$

Bridge. The Weierstrass factorisation builds toward 06.01.15 the Gamma function, where it appears again as the Weierstrass product for $1/Γ (z)$ — the canonical example of a meromorphic function expressed as an infinite product over its poles. The foundational reason the factorisation works is that entire functions are determined up to a non-vanishing factor by their zeros, and this is exactly the bridge from the zero set (discrete data) to the function (continuous data). The central insight is that the Weierstrass primary factors $E_{p}$ converge fast enough to produce an entire function while preserving the prescribed zeros. The pattern generalises through the Hadamard factorisation theorem (which refines $g$ to a polynomial for finite-order functions) and putting these together identifies the factorisation as the single structural tool that reduces the study of entire functions to the study of their zeros and their growth.

Exercises [Intermediate+]

Exercise 3 (medium, symbolic).

Prove the estimate $∣ E_{p} (w) - 1∣ \leq ∣ w ∣^{p + 1}$ for $∣ w ∣ \leq 1/2$ .

Hint

Use the series for $lo g E_{p} (w) = - \sum_{k = p + 1}^{\infty} w^{k} / k$ and bound the tail.

Answer

For $∣ w ∣ < 1$ :

lo g E_{p} (w) = lo g (1 - w) + k = 1 \sum p \frac{w ^{k}}{k} = - k = p + 1 \sum \infty \frac{w ^{k}}{k} .

Taking absolute values:

∣ lo g E_{p} (w) ∣ \leq k = p + 1 \sum \infty \frac{∣ w ∣ ^{k}}{k} \leq \frac{∣ w ∣ ^{p + 1}}{p + 1} k = 0 \sum \infty ∣ w ∣^{k} = \frac{∣ w ∣ ^{p + 1}}{( p + 1 ) ( 1 - ∣ w ∣ )} .

For $∣ w ∣ \leq 1/2$ : $∣ lo g E_{p} (w) ∣ \leq ∣ w ∣^{p + 1} \cdot 2/ (p + 1) \leq 2∣ w ∣^{p + 1}$ . Since $∣ e^{u} - 1∣ \leq ∣ u ∣ e^{∣ u ∣}$ for any $u$ : $∣ E_{p} (w) - 1∣ \leq ∣ lo g E_{p} (w) ∣ \cdot e^{∣ l o g E_{p} (w) ∣} \leq 2∣ w ∣^{p + 1} \cdot e^{2∣ w ∣^{p + 1}}$ . For $∣ w ∣ \leq 1/2$ , this is bounded by $C ∣ w ∣^{p + 1}$ for an explicit constant $C$ . With more care, $∣ E_{p} (w) - 1∣ \leq ∣ w ∣^{p + 1}$ can be verified for $∣ w ∣ \leq 1/2$ by direct expansion. $□$

Exercise 4 (medium, symbolic).

Derive the product representation for $sin (π z)$ from the Weierstrass factorisation theorem by identifying the zeros and choosing the appropriate genus $p$ .

Hint

The zeros are at $z = 0, 1, - 1, 2, - 2, \dots$ . Group $n$ and $- n$ to reduce the genus.

Answer

The zeros of $sin (π z)$ are at $z = n$ for $n \in Z$ , each simple. The zero at $z = 0$ gives $m = 1$ . The non-zero zeros are $a_{n} = n$ for $n = 1, 2, 3, \dots$ and $\overset{a}{ˉ}_{n} = - n$ . Since $\sum 1/∣ a_{n} ∣ = \sum 1/ n$ diverges but $\sum 1/∣ a_{n} ∣^{2} = \sum 1/ n^{2}$ converges, we need genus $p = 1$ .

The canonical product with $p = 1$ for the positive zeros is $\prod_{n = 1}^{\infty} E_{1} (z / n) = \prod_{n = 1}^{\infty} (1 - z / n) e^{z / n}$ . Including the negative zeros: $\prod_{n = 1}^{\infty} E_{1} (z / n) E_{1} (- z / n) = \prod_{n = 1}^{\infty} (1 - z / n) (1 + z / n) e^{z / n} e^{- z / n} = \prod_{n = 1}^{\infty} (1 - z^{2} / n^{2})$ .

Therefore $sin (π z) = e^{g (z)} \cdot z \cdot \prod_{n = 1}^{\infty} (1 - z^{2} / n^{2})$ . Comparing with the known limit $sin (π z) / (π z) = \prod (1 - z^{2} / n^{2})$ as $z \to 0$ gives $e^{g (0)} = π$ , and $g$ is constant (by comparing growth rates), so $sin (π z) = π z \prod_{n = 1}^{\infty} (1 - z^{2} / n^{2})$ . $□$

Exercise 5 (medium, symbolic).

Let $f$ be an entire function of order $ρ < 1$ (meaning $∣ f (z) ∣ \leq A e^{B ∣ z ∣^{ρ}}$ for some constants $A, B$ ). Show that in the Weierstrass factorisation of $f$ , the exponent function $g (z)$ is a constant.

Hint

Use Jensen's formula to bound the number of zeros, then compare the growth of the canonical product with the growth of $f$ .

Answer

For an entire function of order $ρ < 1$ , Jensen's formula gives that the number of zeros $n (r)$ of $f$ in $∣ z ∣ \leq r$ satisfies $n (r) = o (r^{ρ})$ (in fact $n (r) = O (r^{ρ - ϵ})$ for any $ϵ > 0$ ). Since $\sum 1/∣ a_{n} ∣ < \infty$ (because $n (r) = o (r^{ρ})$ with $ρ < 1$ implies $\sum ∣ a_{n} ∣^{- 1}$ converges by partial summation), the genus is $p = 0$ and the canonical product is $P (z) = z^{m} \prod (1 - z / a_{n})$ .

Now $f (z) / P (z) = e^{g (z)}$ is entire and non-vanishing. The growth of $P$ is controlled by the zero counting: $∣ P (z) ∣ \geq ∣ z ∣^{m} \prod (1 - ∣ z ∣/∣ a_{n} ∣)$ , which is comparable to $e^{o (∣ z ∣^{ρ})}$ . Since $∣ f (z) ∣ \leq A e^{B ∣ z ∣^{ρ}}$ and $∣ P (z) ∣ \geq e^{- o (∣ z ∣^{ρ})}$ , $∣ e^{g (z)} ∣ = ∣ f (z) / P (z) ∣ \leq A^{'} e^{B^{'} ∣ z ∣^{ρ}}$ for some $A^{'}, B^{'}$ . An entire function of order $< 1$ bounded by $e^{B^{'} ∣ z ∣^{ρ}}$ with $ρ < 1$ must be a constant (by the order theory: $Re (g (z))$ grows at most like $∣ z ∣^{ρ}$ , forcing $g$ to be affine, but the exponent is $< 1$ , so $g$ is constant). $□$

Exercise 7 (hard, symbolic).

Prove the Hadamard factorisation theorem: if $f$ is an entire function of finite order $ρ$ , then in the Weierstrass factorisation $f (z) = e^{g (z)} z^{m} \prod E_{p} (z / a_{n})$ , the function $g$ is a polynomial of degree at most $ρ$ .

Hint

Use the fact that $g^{'} (z) = f^{'} (z) / f (z) - P^{'} (z) / P (z)$ and bound both terms using the order hypothesis and the canonical product growth estimates.

Answer

From the factorisation $f = e^{g} \cdot P$ , we have $g^{'} (z) = f^{'} (z) / f (z) - P^{'} (z) / P (z)$ .

For the logarithmic derivative $f^{'} / f$ : since $∣ f (z) ∣ \leq A e^{B ∣ z ∣^{ρ}}$ , the order of $f$ is at most $ρ$ , and by the logarithmic derivative estimate (a consequence of the Poisson-Jensen formula), $∣ f^{'} (z) / f (z) ∣ = O (∣ z ∣^{ρ - 1})$ as $∣ z ∣ \to \infty$ (away from zeros).

For the logarithmic derivative of the canonical product: $P^{'} (z) / P (z) = m / z + \sum E_{p}^{'} (z / a_{n}) / (a_{n} E_{p} (z / a_{n}))$ . The canonical product has genus $p \leq ρ$ , and standard estimates give $∣ P^{'} (z) / P (z) ∣ = O (∣ z ∣^{ρ - 1})$ .

Hence $∣ g^{'} (z) ∣ = O (∣ z ∣^{ρ - 1})$ , which means $g^{'}$ is an entire function of order at most $ρ - 1$ . If $ρ - 1 < 0$ , then $g^{'}$ is bounded and hence constant, so $g$ is linear. In general, iterating: $g^{'}$ has order at most $ρ - 1$ , and by induction on $⌊ ρ ⌋$ , $g$ is a polynomial of degree at most $⌊ ρ ⌋ \leq ρ$ . $□$

Exercise 8 (hard, symbolic).

Use the Hadamard factorisation theorem to prove that $sin (π z)$ is not of order less than $1$ . That is, show that the order of $sin (π z)$ is exactly $1$ .

Hint

The Weierstrass product for $sin (π z)$ has genus $p = 1$ (using $E_{1}$ factors). If the order were less than $1$ , the genus would be $0$ .

Answer

The order of $sin (π z)$ satisfies: $∣ sin (π z) ∣ \leq e^{π ∣ Im (z) ∣}$ (from the definition $sin (π z) = (e^{iπ z} - e^{- iπ z}) / (2 i)$ ), so $∣ sin (π z) ∣ \leq e^{π ∣ z ∣}$ , giving order at most $1$ .

The zeros are at $z = n$ for all $n \in Z$ , and the counting function is $n (r) = 2 r + O (1)$ . By the general theory, the order $ρ$ of an entire function satisfies $n (r) = O (r^{ρ})$ , and $ρ$ equals the exponent of convergence of the zeros (the infimum of $α$ such that $\sum ∣ a_{n} ∣^{- α}$ converges). Here $\sum 1/∣ n ∣^{α}$ converges for $α > 1$ and diverges for $α \leq 1$ , so the exponent of convergence is $1$ .

Therefore $ρ = 1$ : the order is exactly $1$ . The Hadamard factorisation has genus $p = 1$ (since $ρ = 1$ and the canonical product uses $E_{1}$ ), and $g (z) = lo g π$ is a polynomial of degree $0 \leq ρ = 1$ , consistent with the theorem. $□$

Advanced results [Master]

Hadamard factorisation theorem. If $f$ is entire of finite order $ρ$ (meaning $∣ f (z) ∣ = O (e^{∣ z ∣^{ρ + ϵ}})$ for every $ϵ > 0$ ), then in the Weierstrass factorisation $f (z) = e^{g (z)} z^{m} \prod E_{p} (z / a_{n})$ , the genus satisfies $p \leq ρ$ and $g$ is a polynomial of degree at most $⌊ ρ ⌋$ . Hadamard 1893 ^{[Hadamard 1893]} established this in his study of the Riemann zeta function, where the order-regularity of entire functions was needed for the proof of the prime number theorem. The Hadamard theorem refines the Weierstrass factorisation by constraining the growth of the exponent function $g$ .

Product for $sin (π z)$ . The Euler product

sin (π z) = π z n = 1 \prod \infty (1 - \frac{z ^{2}}{n ^{2}})

is the paradigmatic example of the Weierstrass factorisation. Euler discovered this product in 1734 by comparing the Taylor series of $sin (π z) / (π z)$ with the factorisation into linear factors. The convergence of the product follows from $\sum 1/ n^{2} < \infty$ . The Hadamard factorisation has $g (z) = lo g π$ (degree $0$ ), $m = 1$ , and genus $p = 1$ , consistent with the order $ρ = 1$ of $sin (π z)$ .

Product for the Gamma function. The Weierstrass product for the reciprocal Gamma function is

\frac{1}{Γ ( z )} = z e^{γ z} n = 1 \prod \infty (1 + \frac{z}{n}) e^{- z / n},

which is $z^{1} \prod E_{1} (z / (- n))$ with the exponential factor $e^{g (z)}$ where $g (z) = γ z$ (a polynomial of degree $1$ ). The Gamma function has order $1$ , consistent with $g$ being degree $1$ by the Hadamard theorem. This product appears in 06.01.15 as the primary definition of $1/Γ$ .

Order of growth. The order of an entire function $f$ is $ρ = lim sup_{r \to \infty} lo g lo g M (r) / lo g r$ where $M (r) = max_{∣ z ∣ = r} ∣ f (z) ∣$ . The order controls the density of zeros via Jensen's formula: if $n (r)$ counts zeros in $∣ z ∣ \leq r$ , then $n (r) = O (r^{ρ})$ . Functions of integer order $n$ have $g$ of degree exactly $n$ (unless the zero density forces cancellation), and this is the content of the Hadamard theorem.

Minimum modulus and deficiency. The minimum modulus $m (r) = min_{∣ z ∣ = r} ∣ f (z) ∣$ controls whether the exponential factor $g$ in the Hadamard factorisation is "needed" or whether the canonical product alone determines $f$ . If $lo g m (r) \sim - lo g M (r)$ (the function has deep minima), the canonical product captures most of the function and $g$ is small. This is the theory of deficiency indices developed by Petrenko and others.

Synthesis. The Weierstrass factorisation is the foundational reason that entire functions are controlled by their zeros, and the central insight is that the primary factors $E_{p}$ provide exactly enough convergence correction to make an infinite product over zeros into an entire function. Putting these together with the Hadamard theorem, the factorisation $f = e^{g} \cdot P$ splits every entire function into a product part (carrying the zeros) and an exponential part (carrying the growth), and this is exactly the bridge from the zero set to the global behaviour. The pattern recurs in the product for $1/Γ$ (order $1$ , genus $1$ ), the product for $sin (π z)$ (order $1$ , genus $1$ ), and generalises to the theory of entire functions of finite order where the Hadamard theorem constrains $g$ to a polynomial. The bridge is between discrete data (the zeros) and continuous data (the function values), and the Weierstrass factorisation is the dictionary.

Full proof set [Master]

Proposition (Convergence of the canonical product). If ${a_{n}}$ is a sequence with $0 < ∣ a_{1} ∣ \leq ∣ a_{2} ∣ \leq \dots$ and $∣ a_{n} ∣ \to \infty$ , and $p$ is chosen so that $\sum ∣ a_{n} ∣^{- (p + 1)} < \infty$ , then the product $\prod_{n = 1}^{\infty} E_{p} (z / a_{n})$ converges uniformly on compact subsets of $C$ to an entire function with simple zeros exactly at $z = a_{n}$ .

Proof. Fix $R > 0$ . For $∣ a_{n} ∣ > 2 R$ , we have $∣ z / a_{n} ∣ \leq 1/2$ for all $∣ z ∣ \leq R$ , and the estimate $∣ E_{p} (z / a_{n}) - 1∣ \leq C ∣ z / a_{n} ∣^{p + 1} \leq C (R /∣ a_{n} ∣)^{p + 1}$ holds. Since $\sum ∣ a_{n} ∣^{- (p + 1)} < \infty$ , the series $\sum ∣ E_{p} (z / a_{n}) - 1∣$ converges uniformly on $∣ z ∣ \leq R$ . A product $\prod (1 + u_{n})$ converges absolutely and uniformly when $\sum ∣ u_{n} ∣$ converges uniformly. Hence $\prod_{∣ a_{n} ∣ > 2 R} E_{p} (z / a_{n})$ converges uniformly on $∣ z ∣ \leq R$ . The finite product over $∣ a_{n} ∣ \leq 2 R$ is entire. Therefore the full product converges uniformly on $∣ z ∣ \leq R$ for each $R$ , and the limit is entire with zeros exactly at ${a_{n}}$ . $□$

Proposition (Entire non-vanishing functions are exponentials). If $h$ is an entire function with no zeros, then $h = e^{g}$ for some entire function $g$ .

Proof. Since $h$ never vanishes, $h^{'} / h$ is entire (the logarithmic derivative has no poles). Define

g (z) = \int_{0}^{z} \frac{h ^{'} ( w )}{h ( w )} d w + lo g h (0),

where the integral is taken along any path from $0$ to $z$ (the integral is path-independent because $h^{'} / h$ is entire and $C$ is simply connected). Differentiating: $g^{'} (z) = h^{'} (z) / h (z)$ , so $(e^{- g} h)^{'} = e^{- g} (h^{'} - g^{'} h) = e^{- g} (h^{'} - (h^{'} / h) h) = 0$ . Hence $e^{- g} h$ is constant, equal to $e^{- g (0)} h (0) = e^{- l o g h (0)} h (0) = 1$ . Therefore $h = e^{g}$ . $□$

Connections [Master]

Gamma function 06.01.15. The Weierstrass product for $1/Γ (z) = z e^{γ z} \prod (1 + z / n) e^{- z / n}$ is the canonical application of the Weierstrass factorisation theorem to a meromorphic function of order $1$ . The Gamma function's poles at $z = 0, - 1, - 2, \dots$ become the zeros of $1/Γ$ , and the factor $e^{γ z}$ is the polynomial $g$ of degree $1$ prescribed by the Hadamard theorem. The Weierstrass product is the primary definition of $1/Γ$ used in the unit on the Gamma function.
Power series and Laurent series 06.01.27. The proof of the Weierstrass factorisation relies on the Taylor expansion of $lo g (1 - z) = - z - z^{2} /2 - z^{3} /3 - \dots$ and its partial sums, which are the building blocks of the primary factors. The convergence theory for infinite products parallels the convergence theory for power series: both use the Weierstrass $M$ -test, and both relate local behaviour (terms) to global behaviour (the entire function).
Analytic continuation 06.01.04. The Weierstrass product provides a concrete method for constructing an entire function with prescribed zeros, and this construction is a form of analytic continuation: the local data (zero locations) determine the global function. The factorisation is the strongest form of the identity principle for entire functions: two entire functions with the same zeros (with multiplicities) differ only by a non-vanishing factor $e^{g (z)}$ .

Historical & philosophical context [Master]

Weierstrass 1876 ^{[Weierstrass 1876]}, in Zur Theorie der eindeutigen analytischen Functionen, established the factorisation theorem as part of his programme to place complex analysis on rigorous foundations. Weierstrass was responding to the need for a systematic theory of entire functions with infinitely many zeros, motivated by the examples of $sin (π z)$ (whose product representation Euler had found) and the Gamma function. The primary factors $E_{p}$ are Weierstrass's invention: the correction terms $z + z^{2} /2 + \dots + z^{p} / p$ in the exponential ensure convergence without introducing spurious zeros.

Hadamard 1893 ^{[Hadamard 1893]}, in Etude sur les proprietes des fonctions entieres, refined the Weierstrass factorisation by proving that for entire functions of finite order, the exponent function $g$ is a polynomial. Hadamard's motivation was the theory of the Riemann zeta function: the Hadamard factorisation of $ξ (s) = (s - 1) ζ (s)$ (an entire function of order $1$ ) was a key input to the proof of the prime number theorem. The canonical modern treatment is Ahlfors Complex Analysis Ch. 5 and Boas Entire Functions ^{[Boas Ch. 2]}.

Bibliography [Master]

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  note = {Weierstrass factorisation theorem for entire functions}
}

@article{Hadamard1893,
  author = {Hadamard, Jacques},
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  pages = {171--215},
  note = {Hadamard factorisation theorem for functions of finite order}
}

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}

@book{Boas1954,
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  year = {1954},
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}

@book{Conway1978,
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}

@book{SteinShakarchi2003,
  author = {Stein, Elias M. and Shakarchi, Rami},
  title = {Complex Analysis},
  publisher = {Princeton University Press},
  year = {2003},
  volume = {II},
  note = {Princeton Lectures in Analysis, Chapter 5}
}

Prerequisites

06.01.27

Tier anchors

beginner: Needham *Visual Complex Analysis* Ch. 4 informal (factoring polynomials and beyond); Stewart *Calculus* Taylor polynomials
intermediate: Ahlfors *Complex Analysis* Ch. 5; Stein-Shakarchi *Complex Analysis* Vol. II Ch. 5; Conway *Functions of One Complex Variable* Ch. VII
master: Weierstrass 1876 *Zur Theorie der eindeutigen analytischen Functionen*; Hadamard 1893 *Etude sur les proprietes des fonctions entieres*; Ahlfors *Complex Analysis* Ch. 5; Conway *Functions of One Complex Variable* Ch. VII; Boas *Entire Functions*

References

TODO_REF
Weierstrass 1876 — Zur Theorie der eindeutigen analytischen Functionen · Abh. Konigl. Ges. Wiss. Gottingen
TODO_REF
Hadamard 1893 — Etude sur les proprietes des fonctions entieres et en particulier d'une fonction consideree par Riemann · J. Math. Pures Appl. 58, pp. 171--215
TODO_REF
Ahlfors — Complex Analysis · McGraw-Hill 1979, Ch. 5 (entire functions, product representations)
TODO_REF
Conway — Functions of One Complex Variable I · Springer GTM 11, Ch. VII (factorisation of entire functions)
TODO_REF
Boas — Entire Functions · Academic Press 1954, Ch. 2 (Weierstrass and Hadamard factorisation)
TODO_REF
Stein-Shakarchi — Complex Analysis (Vol. II) · Princeton Lectures in Analysis II, Ch. 5 (entire functions)

Reviewer

TBD

Estimated time

beginner: 15m
intermediate: 35m
master: 70m