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Consider this function $f(z) = \frac{1}{1+z}$. We can define $f(\infty) = \lim_{z \rightarrow \infty}{f(z)}$, which is zero for this case. Since $f(\frac{1}{t}) \rightarrow \frac{t}{t+1}$ is analytic at point $t=0$, we can deduce that $f(z)$ is analytic at $z=\infty$.

On the other hand, if we consider the type of singularity at $t=0$ of $f(\frac{1}{t})$. There is singularity because of the existence of $\frac{1}{t}$ but the singularity is removable. So the singularity at $z=\infty$ is removable.

The definition is $f(\infty) = \lim_{z \rightarrow \infty}{f(z)}$. So how we discuss the singularity and think of it as removable if the value of $f(\infty)$ is taken from the limit?

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Because the limit exists, you can redefine the function to be equal to the limit, and this is the definition of removable. –  M Turgeon Jun 15 '12 at 15:22
    
I think Strin's point might be confusion over "at $\infty$". For finite numbers, there's a difference between $\lim_{z\rightarrow\ a} f(z)$ and $f(a)$, and removing singularities is about going "from the first to the second", so to speak. But what does it mean for $f(a)$ to exist separately from the limit in the infinite case? Without that it doesn't seem like there's anything to remove. –  Robert Mastragostino Jun 15 '12 at 16:26

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As M Turgeon points out, the whole idea of a removable singularity is that we can "fix" the function at that point so it is equal to the limit, and is then analytic there.

The same idea applies, for example, to the function $f(z)=\dfrac{\sin(z)}{z}$. As it is defined, here, there is a singularity at $z=0$. However, it is removable, since $$\lim_{z\to 0}\frac{\sin z}{z}=1,$$ so we may simply extend the definition of $f$ so that $f(0)=1$, giving us an entire function.

Addendum: Robert Mastragostino brings up a good point, as well. Note that "removable singularity" doesn't (necessarily) mean that we are replacing a given function value with another--indeed, in the example I gave above, $f$ is initially undefined at $z=0$.

An isolated singularity of a function $f$ is a point $z=a$ such that for some $r>0$, we have that $f(z)$ is defined and analytic on $\{z:0<|z-a|<r\}$, but not on $\{z:|z-a|<r\}$. Such a singularity is removable iff there is a function $g$ that agrees with $f$ on $\mathrm{dom}(f)\smallsetminus\{a\}$--note that that doesn't indicate that $a\in\mathrm{dom}(f)$--such that $g(z)$ is defined and analytic on $\{z:|z-a|<r\}$. This may occur by replacing $f(a)$ with $\lim_{z\to a} f(z)$. This may also occur as in the example I gave above, not with replacement, but with extension.

If we are "removing" a singularity at $\infty$, we will generally be extending, rather than replacing a pre-existing function value. A notable exception is with linear fractional transformations $T(z)=\dfrac{az+b}{cz+d}$, where $ad-bc\neq 0$. In such a case, we tend to explicitly define $T(\infty)=\frac{a}{c}$ at the outset.

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But if the value of $f(\infty)$ is extended and defined as $T(\infty) = \frac{a}{c}$, then the function is also analytic. Can we say the point at infinity is both "removable" and analytic. –  Strin Jun 16 '12 at 14:51
    
True. If we extend it so that $T(\infty)=a/c$, then the function is indeed analytic at $\infty$. If we didn't extend it, then we would have a removable singularity, but the convention is to go ahead and extend it. We also tend to define $T(-d/c)=\infty$, but in that case, we are dealing with a pole, and so the map is not analytic there (though it is continuous, in a sense). –  Cameron Buie Jun 16 '12 at 15:31

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