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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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Thanks! I'm reading something that says $f(z)=g(1/z)=z^nh(1/z)$ where $h(1/z)$ is bounded near $z=\infty$. It then says from this that for some $M$, $|f(z)|\leq M|z|^n$, which seems to follow from $|h(1/z)|\leq M$ outside a large enough circle. How can this conclusion be seen, at least intuitively? Does being bounded near $\infty$ always imply that a function is bounded by some constant if we get sufficiently far out? – MJY Mar 22 '12 at 11:28
@MJY: Just as you say, $h(1/z)$ is bounded by (say) $M$ outside of a large enough circle, so $|f(z)|=|z|^n|h(1/z)|\le M|z|^n$ outside it as well. (Presumably $f$ is also holomorphic and thus bounded inside the circle as well; if you put the two bounds together...) – anon Mar 22 '12 at 11:33
sorry, I see why $|f(z)|$ would be bounded, but I don't see why $h(1/z)$ being bounded at $z=\infty$ means it is bounded by $M$ outside a large enough circle. Unless that's just what the statement means by definition... – MJY Mar 22 '12 at 11:34
@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. – anon Mar 22 '12 at 11:36

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