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I don't know how you want to define "pathological", but some of the corollaries of Runge's theorem give you functions with interesting properties. One of mine: there is a rational function $f$ such …

No. Consider $$f(z) = \sum_{n=1}^\infty \frac{z^{2^n}}{n^2}$$ By the Ostrowski-Hadamard gap theorem, the natural boundary is the unit circle. But the series converges absolutely on the unit circle, …

The series $$f(z) = \sum_{n=1}^\infty \dfrac{z^{2^n}}{n}$$ converges almost everywhere on the unit circle by Carleson's theorem (it is the Fourier series of an $L^2$ function). However, it diverges o …

The (physicists') Hermite polynomials are $$ H_n(x) = (-1)^n e^{x^2} D^n e^{-x^2}$$ And their roots are real. For that you don't need to know they are Hermite polynomials: just Rolle's theorem. Se …

Here is an animation of the zeros of the first $100$ Bernoulli polynomials, produced using Maple. For the number of real roots, see OEIS sequence A094937 and references there. EDIT: As requested b …

For (1), just take $f(z) = z$. (Or do you want to call that "converging to $\infty$"?) For (2), the problem is what happens on the imaginary axis. Whether $f(it)$ diverges or converges to $0$ as $t …

For a counterexample, let $\gamma$ be the unit circle. Let $$A_n = \{z \in \gamma:\; \text{Im}(z) \in [-1,0] \cup [1/n, 1]\}$$ By Runge's theorem there is a polynomial $f_n$ such that $|f_n| < 1/n$ …

For any polynomial $f$ of degree $n$, $$f(\phi(x)) - f(x) = (\phi(x) - x) g(\phi(x),x)$$ where $g$ is a bivariate polynomial of degree $n-1$, so the symmetries will be $\phi(x) = x$ and the roots of …

Some cases in point: $A_1 = B_1 = \exp(z)$ entire with no zeros, $A_1 \star B_1 = I_0(2 \sqrt{z})$ with infinitely many. $$\eqalign{A_2 &= \sum_{n=0}^\infty \frac{z^n}{n!!} = 1 + z e^{z^2/2} + \sqrt …

On the contrary: the Jacobi theta function $$\theta_2(0,q) = 2 q^{1/4}\sum_{n=0}^\infty q^{n(n+1)}$$ has a natural boundary at $|q|=1$. Your function is $f(t) = (1/2) \theta_2(0,\exp(-t))$, so you ca …

$f$ has an entire extension iff the radius of convergence of the Maclaurin series of $f$ is $+\infty$, i.e. iff $\limsup_{n \to \infty} |a_n|^{1/n} = 0$ where $a_n$ are the coefficients of that series …

By Hadamard's theorem, a lacunary series $\sum_k c_k z^{\lambda_k}$ with finite radius of convergence where $\inf_k \lambda_{k+1}/\lambda_k > 1$ can't be analytically continued outside its circle of c …

Yes, if $Z$ has Hausdorff 1-dimensional measure $0$. Then for any $\epsilon > 0$ you can cover $Z$ by a finite number of disks the sum of whose circumferences is less than $\epsilon$. The integral of …

If $\Re(s) > 1/2+\epsilon$ for $\epsilon > 0$ we have $\log(1+p^{-s}) = \sum_{j=1}^\infty p^{-sj}/j$ so $|f_p(s)| \le \sum_{j=2}^\infty p^{-\Re(s)j}/j \le p^{-1-2\epsilon}/(1-p^{-1/2-\epsilon})$. S …

No, e.g. you may run into a singularity. For example, take $X = {\mathbb C}$, $u(t) = t$, $a=0$ and $\phi(z) = \frac{1}{1-2z}$ in a neighbourhood of 0. The pole at $t = 1/2$ stops the analytic conti …