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Suppose $f(z)$ is some analytic function which is bounded near $0$. Then $f(1/z)$ is bounded near $\infty$. What exactly does that last statement mean practically?

Does it mean $|f(1/z)|$ is bounded somehow?

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It means precisely the first statement: a function $g(z)$ bounded "at $z=\infty$" when $g(1/z)$ is bounded in a neighborhood of the origin $z=0$. Thus to say $f(1/z)$ is bounded at infinity is to say that $f(z)$ is bounded at zero. One also says functions are holomorphic at infinity, et cetera.

To understand this intuitively, just picture the point infinity located on the Riemann sphere; looking at a neighborhood of this point and examining a function $f$ there is equivalent to flipping the sphere upside down (done by taking the reciprocal) and examining $f(1/z)$ at the origin ($1/z$ at the origin corresponds to $\infty$ on the sphere).

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    @MJY: The intuition here is that any loop going around infinity on the Riemann sphere is also a loop in the complex plane; the side of the loop where $\infty$ is located, on the sphere, will be "outside" the loop when viewed in the plane. This is why any neighborhood of infinity is "outside a large enough circle" in the complex plane.2012-03-22