Is $g(z)=\frac{1}{z}+\frac{1}{z^2+1}+\frac{1}{(z^2+1)^2 +1}+...$ analytic for $|z|>2$?

Question

Let $z$ be a complex number. Let |.| denote be the absolute value. Let $n$ be a positive integer. Let $f_1(z)=z^2+1$. Let $f_n(z)=f_1(f_{n-1}(z)).$

Is $g(z)=\dfrac{1}{z}+\dfrac{1}{f_1(z)}+\dfrac{1}{f_2(z)}+$ $...$ analytic for $|z|>2$ ?

Answer 1

The function is analytic on $|z|>2$ if the denominators of every term are nonzero for $|z|>2$. So we want to determine if there exists a solution for $f_n(z)=0$ when $|z|>2$ for any $n$.

Now, it's obvious that both $z$ and $z^2+1$ have zeros only within $|z|<2$. Let's look at the rest of the sequence.

If $f_n(z)=0$, then we know that $f_{n-1}(z)$ must be a zero of $z^2+1$ - that is, $f_{n-1}(z)=\pm i$. And it iterates - for some $h$ that is a root of $f_{m}(h)=0$, $f_{n-m}(z)=h$ when $f_n(z)=0$.

And so, the easiest way to look at this is "if $|h|\leq 2$, can $|z|>2$?"

Suppose that $|h|\leq 2$ and is a solution to $f_{n-1}(z)=0$. Then we have that

$$ f_1(z)=z^2+1=h=h_x+ih_y $$ where $h_x^2+h_y^2\leq 4$. And so we have

$$\begin{align} z &= \pm\sqrt{(h_x-1)+ih_y}\\ & = \pm\sqrt{\sqrt{(h_x-1)^2+h_y^2}e^{i\tan^{-1}\left(h_y/(h_x-1)\right)}}\\ &= \pm \sqrt[4]{(h_x-z)^2+h_y^2}e^{i\tan^{-1}\left(h_y/(h_x-1)\right)/2} \end{align}$$ And so,

$$\begin{align} |z| &= \sqrt[4]{(h_x-1)^2+h_y^2}&\\ & = \sqrt[4]{h_x^2+h_y^2-2h_x+1}&\\ &\leq \sqrt[4]{4-2h_x+1} &\textrm{ as } &h_x^2+h_y^2\leq 4\\ &\leq \sqrt[4]{9} &\textrm{ as } &-2\leq h_x\leq2\\ &= \sqrt{3}\\ &<2 \end{align}$$ Therefore, $|z|<2$ if $|h|\leq 2$, and there cannot be any solution to $f_n(z)=0$ on $|z|>2$ for any $n$. Therefore, the function is analytic.

Answer 2

Hint: Note that $$ g(1/z)=z+\frac{z^2}{1+z^2}+\frac{z^4}{(1+z^2)^2+z^4}+\dots $$ where each successive term is the reciprocal of ($1$ plus the square of the reciprocal of the previous term). The numerator of each term is $z^{2^k}$ and the denominator is a degree $2^k$ polynomial with constant coefficient $1$.

Hint: Note that if $|z|\gt2$ then $|z^2+1|\gt\frac32|z|$