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I don't know what kind of condition you're looking for, but whatever condition it is must essentially work through the Argument Principle: as $z$ goes around the (positively oriented) curve $C$, it mu …

If $w = x + i y$, the curve $|f(w)|^2=1$ can be written as $Q(x,y) = 1$ where $$ Q(x,y) = \dfrac{{x}^{6}}{36}+\dfrac{{x}^{4}{y}^{2}}{12}+\dfrac{{x}^{2}{y}^{4}}{12}+\dfrac{{y}^{6}}{36}+ \dfrac{{x}^{4}} …

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 (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 …

One sufficient condition is that $k > 0$ and $\text{Re}(a) - |b| - |c| > 0$.

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 …

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 …

Try $P(z) = z^2 + z + 1$. Its zeros are in the sector $-2\pi/3 \le \theta \le 2\pi/3$, but the zero of $P'$ is $-1/2$ which is not.

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. 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, …

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 …

$\Delta(z)$ does not have a nonpositive real eigenvalue for $|z| < 4$, so the principal branch of the logarithm is defined and analytic on a neighbourhood of its spectrum, and thus the holomorphic fu …

There are some easy bounds. Let $f(z) = \dfrac{g_1(z) g_2(z)}{h_1(z) h_2(z)}$ where $g_1, g_2, h_1, h_2$ are polynomials not identically $0$, $g_1$ and $h_1$ having their roots inside the circle $C$ …

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 …

Yes: if there is a conformal map from $\Omega$ to $A(r)$ that takes $C_1$ to $\{|z|=r\}$ and $C_2$ to $\{|z|=1\}$, then there is a Möbius transformation that does so, and the pole of this function is …