I was thinking a problem like this:

Imagine you have an entire function $f(z)$. Then u can write $f(z)=\sum_{n\ge0}a_nz^n$ for Taylor; Wich is a serie. Then i consider $\lim_{z \to \infty}f(z)$. And now i have multiple questions:

a) Does it always exists?

b) If there exists an $n_0$ that $a_n=0$ $\forall n\ge n_0$. It exists too?

I have a result wich say that if we have an entire function and $\lim_{z \to \infty}f(z)=\infty$ then necessary it's a polynominal. I just want to understand theory of complex funcionts limits.


First of all, let us be clear: the definition of limit at $\infty$ that I work with in the context of complex analysis is this:

  • $\lim_{z\to\infty}f(z)=w\in\mathbb C$ if $(\forall\varepsilon>0)(\exists M\in\mathbb{R}^+):|z|>M\implies|f(z)-w|<\varepsilon$;
  • $\lim_{z\to\infty}f(z)=\infty$ if $(\forall M\in\mathbb{R}^+)(\exists N\in\mathbb{R}^+):|z|>M\implies|f(z)|>N$.

The limit $\lim_{z\to\infty}f(z)$ doesn't have to exist. It doesn't exist if $f$ is, say, the exponential function, the sine function or the cosine function. On the other and, if $a_n=0$ if $n$ is large enough, then the limit exists, and it is equal to $\infty$. Finally, this works in the opposite direction too: if $f$ is entire and $\lim_{z\to\infty}f(z)=\infty$, then $f$ is polynomial. That's a consequence of the Casoratti-Weierstrass polynomial.

  • $\begingroup$ Okey thanks. One question:How can I prove that if $f$ is a polynominal then this limit is equal to $\infty$. $\endgroup$ – Lecter Jun 27 '18 at 16:11
  • $\begingroup$ One example: If i have $f(z)=z^2$, then the limit to $\infty$ doesn't exists because if i write $z=x$, i got $+\infty$, but if i write $z=iy$, i got $-\infty$. Doesn't it is a contradiction? $\endgroup$ – Lecter Jun 27 '18 at 16:18
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    $\begingroup$ I explained What I meant by limits with $\infty$ in the context of complex analysis. There's no $+\infty$ or $-\infty$ here. There's just $\infty$. $\endgroup$ – José Carlos Santos Jun 27 '18 at 16:22
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    $\begingroup$ If $f(z)=a_0+a_1z+\cdots+a_nz^n$ with $a_n\neq0$, then$$\bigl|f(z)\bigr|\geqslant|a_n|.|z|^n\left(1-\left(\frac{|a_{n-1}|}{|z|}+\cdots+\frac{|a_0|}{|z|^n}\right)\right)$$and therefore $\lim_{z\to\infty}f(z)=\infty$. $\endgroup$ – José Carlos Santos Jun 27 '18 at 16:25

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