Is the absolute value of the j-invariant bounded from below on an annulus

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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The $j$-invariant is meromorphic with a simple pole at $\infty$. In other words, $1/|j|$ is bounded on each $B(\epsilon)$. –  Kevin Ventullo Oct 9 '11 at 19:31
@Kevin: he is asking if it is bounded BELOW. –  Igor Rivin Oct 9 '11 at 19:44
$1/|j|$ 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. –  Noam D. Elkies Oct 9 '11 at 19:48
@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. –  GH from MO Oct 9 '11 at 19:56
@GH: you're right; not sure what I was thinking. –  Noam D. Elkies Oct 9 '11 at 21:58