I often run into problems when i consider entire functions as Some sort of " polynomial of infinite degree ". Yet I do not know why most of the time.
It seems intuitive. Apart from the exp function not having a zero ofcourse.
Here is an example.
Let $f(z)$ be an analytic function such that $| f(0) | > 0$ and $| f(1)| > 1$.
Now consider the arguments below that give a false proof.
Let $n$ be the " degree " of the polynomial of infinite degree. We take the limit as n Goes to infinity.
Let us try to find ( complex) solutions $r$ such that
$$|r| > 1$$
$$ f(r) = 0 $$
We argue as follows :
Thus, let us assume $|r|\ge 1$. Since $$ f(r) = a_0 + a_1r + \cdots + a_{n-1}r^{n-1} + r^n =0 $$ then $$ r = -\frac{a_0}{r^{n-1}} - \frac{a_1}{r^{n-2}} - \cdots - a_{n-1}. $$ By taking absolute values and using the triangular inequality, we obtain $$ |r| \le \frac{a_0}{|r|^{n-1}} + \frac{a_1}{|r|^{n-2}} + \cdots + |a_{n-1}|. $$ Finally, using that $|r|\ge 1$ (so that $\frac{1}{|r|}\le 1$), we obtain the required upper bound $$ |r| \le |a_0| + \cdots + |a_{n-1}|. $$
Since when $ n$ Goes to infinity then $n-1$ Goes to infinity as well and
$$ \infty >> |a_0| + \cdots + |a_{\infty}| >= |f(1)| > 1$$ $$| r | \le |a_0| + \cdots << \infty $$
But this would imply there is boundary on the zero's of $f(z)$. In fact in the case when $f(z)$ has infinitely many zero's they all belong within an absolute value ; there making the function no longer analytic !!
This is clearly a wrong proof.
But why ???
You have started by assuming $r$ is a zero of the polynomial $$g_n(z)=a_0 + a_1z + \cdots + a_{n-1}z^{n-1} + z^n.$$ You prove some statements that are true about any such root. Then you "take the limit as $n\to\infty$", and seem to be concluding that similar statements must be true of any zero of the function $f(z)=\sum_{k=0}^\infty a_kz^k$. This doesn't follow at all though. If $r$ is a zero of $f(z)$, then it is probably not a zero of $g_n(z)$ for any $n$ (let alone for all $n$), so what you have proven about zeroes of $g_n(z)$ does not tell you anything about $r$ (at least not in any obvious way).
Moreover, it is not even true that $g_n(z)$ converges to $f(z)$ as $n\to\infty$. What is true is that $f_n(z)=a_0 + a_1z + \cdots + a_{n-1}z^{n-1}$ converges to $f(z)$, but $g_n(z)=f_n(z)+z^n$ will be very far from $f_n(z)$ if $|z|>1$ and $n$ is large. So even as a nonrigorous heuristic argument, you are doing the wrong calculation.