Chapter 4

Which of the following products are absolutely convergent? Find the corresponding values, when they exist. (a) \(\prod_{\nu=2}^{\infty}\left(1-\frac{1}{\nu}\right)\), (b) \(\prod_{\nu=2}^{\infty}\left(1-\frac{1}{\nu^{2}}\right)\), (c) \(\prod_{\nu=2}^{\infty}\left(1-\frac{2}{\nu(\nu+1)}\right)\), (d) \(\prod_{\nu=2}^{\infty}\left(1-\frac{2}{\nu^{3}+1}\right) .\)

Prove the WE1ERSTRASS Product Theorem by means of MITTAG-LEFFLER's Theorem, by first considering the principal part distribution $$ \left\\{\frac{m_{n}}{z-s_{n}} ; n \in \mathbb{N}\right\\} $$ and then observing that \(f\) is a solution for the zero distribution \(\left\\{\left(s_{n}, m_{n}\right) ;\right.\) \(n \in \mathbb{N}\\}\), iff \(\frac{f^{\prime}}{f}\) is a solution for the principal part distribution \(\left\\{\frac{m_{n}}{z-s_{n}} ; n \in \mathbb{N}\right\\}\)

Let \(D=\\{z \in \mathbb{C} ; \quad|z|>1\\}\). Can there exist any conformal map from \(D\) onto the punctured nlane \(C^{*} ?\)

The product \(\prod_{\nu=0}^{\infty}\left(1+z^{2^{\nu}}\right)\) is absolutely convergent, \(i f f|z|<1 .\) If this is the case, then $$ \prod_{\nu=0}^{\infty}\left(1+z^{2^{\nu}}\right)=\frac{1}{1-z} $$

The two annuli $$ r_{\nu}<|z|

Show that the sequence \(\left(\gamma_{n}\right)\) defined by $$ \gamma_{n}:=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\log n $$ is (strictly) decreasing, and bounded from below by \(0 .\) Hence the following limit exists: $$ \gamma:=\lim _{n \rightarrow \infty} \gamma_{n} \approx 0.577215664901532860606512090082402431042159 \ldots $$ (The Euler-Mascheroni Constant).

Show: $$ \frac{\pi}{\cos \pi z}=4 \sum_{n=0}^{\infty} \frac{(-1)^{n}(2 n+1)}{(2 n+1)^{2}-4 z^{2}} $$ and derive from this $$ \frac{\pi}{4}=\sum_{n=0}^{\infty}(-1)^{n} \frac{1}{2 n+1}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots $$

Show that the EuLER Product Formula for \(1 / \Gamma\) can be obtained from GAuss' representation of \(\Gamma\), and conversely. For this, recall: $$ \begin{aligned} &\Gamma(z)=\lim _{n \rightarrow \infty} \frac{n ! n^{2}}{z(z+1) \ldots(z+n)} \\\ &\frac{1}{\Gamma(z)}=z e^{\gamma=} \prod_{n=1}^{\infty}\left(1+\frac{z}{n}\right) e^{-x / n} \end{aligned} $$ (C.F. GAUSS)

Let \(f: \mathbb{C} \rightarrow \mathbb{C}\) be a meromorphic function, such that all its poles are simple with integral residues. Then there exists a meromorphic function \(h: \mathbb{C} \longrightarrow \mathbb{C}\) with \(f(z)=h^{\prime}(z) / h(z)\).

Show for \(z \in \mathbb{C} \backslash S, S:=\\{0,-1,-2,-3, \ldots\\}\) $$ \lim _{n \rightarrow \infty} \frac{\Gamma(z+n)}{n^{2} \Gamma(n)}=1 $$

Determine the image of $$ D=\\{z \in \mathbb{C} ; \quad|\operatorname{Re} z||\operatorname{Im} z|>1,0<\operatorname{Re} z, \operatorname{Im} z\\} $$ under the map \(\varphi(z)=z^{2}\).

Show for \(z \in \mathbb{C} \backslash S, S:=\\{0,-1,-2,-3, \ldots\\}\) $$ \lim _{n \rightarrow \infty} \frac{\Gamma(z+n)}{n^{2} \Gamma(n)}=1 $$

Prove the following uniqueness result (H. POINCARÉ, 1884\()\) : If \(D \subset \mathbb{C}\) is an elementary domain which is not \(\mathbb{C}\), and if \(z_{0} \in D\) is a fixed point in \(D\), then there exist exactly one conformal map $$ \varphi: D \longrightarrow \mathbb{E} \quad \text { with } \varphi\left(z_{0}\right)=0 \text { and } \varphi^{\prime}\left(z_{0}\right)>0 $$

Let \(D \subset \mathbb{C}\) be an elementary domain, and let \(f: D \rightarrow \mathbb{E}\) be a conformal map. If \(\left(z_{n}\right)\) is a sequence in \(D\) with \(\lim _{n \rightarrow \infty} z_{n}=r \in \partial D\), then the sequence \(\left(\left|f\left(z_{n}\right)\right|\right)\) converges to 1 . Give an example of a sequence \(\left(z_{n}\right)\) converging to a boundary point of \(D\), such that the image sequence \(\left(f\left(z_{n}\right)\right)\) by a conformal map \(f: D \rightarrow \mathbb{E}\) does not converge to a boundary point of \(\mathbb{E}\).

For \(\alpha \in \mathbb{C}\), and \(n \in \mathbb{N}\) let $$ \left(\begin{array}{l} \alpha \\ n \end{array}\right):=\frac{\alpha(\alpha-1) \cdots(\alpha-n-1)}{n !}, \quad\left(\begin{array}{l} \alpha \\ 0 \end{array}\right):=1 $$ Show that for all \(\alpha \in \mathbb{C} \backslash \mathbb{N}_{0}\) $$ \left(\begin{array}{l} \alpha \\ n \end{array}\right)=\frac{(-1)^{n} \Gamma(n-\alpha)}{\Gamma(-\alpha) \Gamma(n+1)} \sim \frac{(-1)^{n}}{\Gamma(-\alpha)} n^{-\alpha-1} \quad \text { for } n \rightarrow \infty $$ i.e. the quotient of the expressions on both sides converges to 1 for \(n \rightarrow \infty\)