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Here's a problem I was just working on:

Let $f$ have an essential singularity at $0$. Show that there is a sequence of points $z_n \to 0$ such that $z_n^n f(z_n)$ tends to infinity.

I know already that there exists a sequence $z_n \to 0$ such that $f(z_n)$ tends to any complex number I want, hence I can get a sequence that tends to infinity. The problem is that I need this sequence to tend to infinity really fast.

What I did so far was look at the function $g_n(z) := z^n f(z)$, $n \geq 1$. This obviously also has an essential singularity at $0$, so I can find a sequence $z_{n1}, z_{n2}, ...$ such that $\lim\limits_k g_n(z_{nk}) = \infty$. Do you think it's possible to extract from the array $z_{nk}, (n,k) \in \mathbb{N}^2$ a subsequence $z_l$ such that $g_l(w_l) \to \infty$ as $l \to \infty$? I tried for awhile but I'm just not very good at these kinds of arguments.

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Yes. It is possible, but I think it will be easier to rebuild it from scratch than to extract such a sequence.

You can define $z_n$ in the following way:

For each $n$, there exists $z_n \in B(0,1/n)$ such that $|g_n(z_n)| > n$ (since, as you said yourself, there is a sequence that tends to zero and takes $g_n$ to $\infty$). This is your sequence. It should be easy to finish the proof from here.

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  • $\begingroup$ Thanks for answering. I'm sorry but I just realized I typed the question wrong. I've edited it. $\endgroup$ – D_S Jul 18 '15 at 23:24
  • $\begingroup$ Re-edited my answer then :) $\endgroup$ – amirbd89 Jul 18 '15 at 23:33
  • $\begingroup$ $z_n = f(u)$, where $|f(u)| = \max_{|z| = 1/n} |f(z)| $ ? $\endgroup$ – reuns Jul 20 '15 at 11:44
  • $\begingroup$ Hmmm... I think that you meant is $z_n = u$. I'm not sure that this will work, even though I can't fin a counter example. But you know that $sup_{|z| < 1/n} |g_n(z)| = \infty$ you can find $z_n \in B(0, 1/n)$ such that $g_n(z_n) > n$. $\endgroup$ – amirbd89 Jul 21 '15 at 7:35
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You should be able to extract such a sequence from $g_n(z_{n,k})$. To put this sequence property slightly differently, you know that for any $n \in \mathbb{N}$, any complex number $w$, and any $\varepsilon > 0$, you can find a complex number $z$ such that $|g_n(z) - w| < \varepsilon$. So, fix $n \in \mathbb{N}$, choose $\varepsilon = 1$, and $w = n$, and you get a sequence of corresponding $z_n$'s satisfying, $$|g_n(z_n) - n| < 1.$$ Basically, $g_n(z_n)$ is a sequence that's always within distance $1$ of $n$, so it shouldn't tend to anything other than $\infty$. To prove this, use reverse triangle inequality. We have, $$||g_n(z_n)| - n| \le |g_n(z_n) - n| < 1,$$ so $$-1 + n < |g_n(z_n)| < 1 + n.$$ By the squeeze theorem, $|g_n(z_n)| \rightarrow \infty$.

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