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It is not quite clear what you mean by calculate. This limit is a familiar, classical object: it is called the Green function of the complement of the Julia set, or the equilibrium potential of the Ju …

There is no such example. Let $(f_n)$ be a sequence of characteristic functions of probability measures $\mu_n$ which converges a. e. to a characteristic function $f$ of a probability measure $\mu$. Y …

This has nothing to do with complex numbers, or Rouche's theorem. In any normed field, denote $r=\max|r_j|$ and $A=\max|a_j|$. Then $$r^n\leq An\max\{1,r^{n-1}\}.$$ So if $r\leq 1$ then $r\leq (An)^{1 …

The assumption of the Lemma is that $f$ has a finite limit as $z\to \zeta_0$. This assumption does not hold in any of the two examples that you mention. In these examples, $f$ has a limit only when $z …

Cosine has one attracting fixed point $a\approx0.7390851$, and both critical values $\pm1$ are attracted to it. Then it follows from general theorems of dynamics of entire functions that it has one co …

If $0$ is a pole, then there is a formula for the residue $$\mathrm{res}_0=\lim_{x\to 0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}x^nf(x),$$ which involves only the restriction of $f$ on the real line. …

More detail on weak convergence of subharmonic functions, including a proof of this statement, can be found in his other book: Hormander, Notions of convexity, Theorems 3.2.12 and 3.2.13.

In general there is no continuous function between u.s.c. $f$ and l.s.c. $g\leq f$. For example, take $f(x)=1, 0\leq x\leq 1;\; f(x)=0, 1<x\leq 2$, this is u.s.c. Now $g(x)=1, 0\leq x<1;\; g(x)=0, 1\l …

Of course, you cannot put $\epsilon=0$ under the integral. You must evaluate it for $\epsilon\neq 0$ and then pass to the limit. The correct way to find this integral is to spread it from $-\pi/2$ to …

The region of convergence is the union of polydisks $\{(x,y):|x|<r_1,|y|<r_2\}$ such that $$\limsup_{m+n\to\infty}|c_{m,n}r_1^nr_2^m|^{1/(m+n)}\leq 1,$$ which is the generalization of the Cauchy-Hadam …