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Let $j:\mathbf{H}\to \mathbf{C}$ be the $j$-invariant. It's a modular function for $\Gamma(1) = \textrm{PSL}_2(\mathbf{Z})$.

For $\epsilon>0$ small, let $B(\epsilon)$ be the image of the strip $$\{x+iy: 0\leq x <1 , y> \frac{1}{\epsilon}\}\subset \mathbf{H}$$ under the quotient map $\mathbf{H}\longrightarrow X(1)$. Note that $B(\epsilon)$ is an open disc around the cusp $\infty$.

Is the absolute value of the $j$-invariant bounded from below (by a positive real number depending on $\epsilon$) on $B(\epsilon)$?

The answer to this is probably no. But what if we take an annulus?

Is the absolute value of the $j$-invariant bounded from below on $B(\epsilon) - B(\epsilon/2)$ in terms of $\epsilon$?

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    $\begingroup$ The $j$-invariant is meromorphic with a simple pole at $\infty$. In other words, $1/|j|$ is bounded on each $B(\epsilon)$. $\endgroup$
    – Kevin Ventullo
    Commented Oct 9, 2011 at 19:31
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    $\begingroup$ @Kevin: he is asking if it is bounded BELOW. $\endgroup$
    – Igor Rivin
    Commented Oct 9, 2011 at 19:44
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    $\begingroup$ $1/|j|$ is indeed bounded on $B(\epsilon)$ as K.Ventullo notes, but the question concerns lower bounds on $|j|$. The answer is still yes (even without excluding $B(\epsilon/2)$), because $j(\tau) = 1/q + O(1)$ as $q = \exp 2\pi i \tau \rightarrow 0$, and $1/|q| = \exp 2 \pi y > \exp(2 \pi / \epsilon) $ is very large for $\epsilon$ small. $\endgroup$
    – Noam D. Elkies
    Commented Oct 9, 2011 at 19:48
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    $\begingroup$ @Igor and Noam: $1/|j|$ being bounded means that $|j|$ is bounded from below. So Kevin's answer was perfect unless I am missing something. $\endgroup$
    – GH from MO
    Commented Oct 9, 2011 at 19:56
  • $\begingroup$ @GH: you're right; not sure what I was thinking. $\endgroup$
    – Noam D. Elkies
    Commented Oct 9, 2011 at 21:58

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