# Is the function really analytic at $\infty$?

I want to verify whether the complex-valued function $f(z)= \dfrac{1}{z}+\dfrac{1}{z^5}$ is analytic at $\infty$.

I attempted to do this by approximating the function by its power series expansion $$\dfrac{1}{z}+\dfrac{0}{z^2}+\dfrac{0}{z^3}+\dfrac{0}{z^4}+\dfrac{1}{z^5}+\frac{0}{z^6}+\frac{0}{z^7}+\ldots.$$ Since the function is expressed in this power series, I would say that it is analytic at $\infty$ with a simple zero at $\infty$. What if I considered the following function: $f(z)=\dfrac{z^2 +1}{z^2 -1}$? I was able to figure out that it is also analytic at $\infty$ through use of the power series expansion. However, does this function have any zeros of some particular order at $\infty$? In the series, my constant term $a_0=1$, so I would think not, but I'm not sure if I'm correct. I would appreciate any expert input, thanks.

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What is your definition of a function being analytic at $\infty$? – Zhen Lin Oct 30 '11 at 17:37

Let $f:\Omega\to\mathbb{C}$ be a holomorphic function where $\Omega$ is an open subset of $\mathbb{C}$ such that $\{\frac{1}{z}:z\in\Omega\}$ contains a deleted neighborhood of $0$ in $\mathbb{C}$. I believe that $f:\Omega\to\mathbb{C}$ has a removable singularity at $\infty$ if and only if $z\to f\left(\frac{1}{z}\right)$ has a removable singularity at $0$. (Exercise!)

Let $f$ be the function in the first line of your question. We can write $f\left(\frac{1}{z}\right)=z+z^5$ for all $z\in\mathbb{C}\setminus \{0\}$. In particular, it is clear that $z\to f\left(\frac{1}{z}\right)$ has a removable singularity at $0$.

Let $f$ be the function in the fifth line of your question. We can write $f\left(\frac{1}{z}\right)=\frac{1+z^2}{1-z^2}$ for all $z\in\mathbb{C}\setminus \{0\}$. In particular, it is clear that $z\to f\left(\frac{1}{z}\right)$ has a removable singularity at $0$.

However, in this case $\lim_{z\to 0} f\left(\frac{1}{z}\right) = 1$. In particular, $f$ does not have a zero at $\infty$.

The following exercises are relevant to your question:

Exercise 1: Let $f:\mathbb{C}\to\mathbb{C}$ be an entire function. If $f$ has a removable singularity at $\infty$, then prove that $f$ is contant. (Hint: Liouville's theorem.)

Exercise 2: Let $f:\Omega\to\mathbb{C}$ be a holomorphic function where $\Omega$ is an open subset of $\mathbb{C}$ such that $\{\frac{1}{z}:z\in\Omega\}$ contains a deleted neighborhood of $0$ in $\mathbb{C}$. We write that $f$ has a pole of order $m$ at $\infty$ if the holomorphic function $z\to f\left(\frac{1}{z}\right)$ (defined in a deleted neighborhood of $0$) has a pole of order $m$ at $0$. Prove that $f$ has a pole of order $m$ at $\infty$ if and only if there are complex numbers $c_1,\dots,c_m$ with $c_m\neq 0$ such that $f(z) - \sum_{k=1}^{m} c_kz^k$ has a removable singularity at $\infty$.

Exercise 3: Let $f:\mathbb{C}\to\mathbb{C}$ be an entire function. If $f$ has a pole of order $m$ at $\infty$ (for some positive integer $m$), then prove that $f$ is a polynomial of degree $m$.

Exercise 4: Let $f:\Omega\to\mathbb{C}$ be a holomorphic function where $\Omega$ is an open subset of $\mathbb{C}$ such that $\{\frac{1}{z}:z\in\Omega\}$ contains a deleted neighborhood of $0$ in $\mathbb{C}$. Define what it means for $f$ to have an essential singularity at $\infty$.

Exercise 5: Prove that the entire function $f(z)=e^z$ has an essential singularity at $\infty$.

I hope this helps!

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A quick question, for the function in the first part of my question, is $f(\frac{1}{z})= z+ z^5$, or isn't it actually supposed to be $f(\frac{1}{z})= \frac{z^5}{1+ z^4}$? Please clarify, thanks. – Libertron Oct 29 '11 at 23:25
@Sachin Dear Sachin, you are correct that $f\left(\frac{1}{z}\right)=z+z^5$ for all $z\in\mathbb{C}\setminus \{0\}$. Please remember that there is a difference between $f\left(\frac{1}{z}\right)$ and $\frac{1}{f(z)}$ for $z\in\mathbb{C}$. Indeed, $f\left(\frac{1}{z}\right) = z+z^5$ for all $z\in\mathbb{C}\setminus \{0\}$ but $\frac{1}{f(z)}=\frac{z^5}{1+z^4}$ for all $z\in\mathbb{C}\setminus \{0\}$. Regards, – Amitesh Datta Oct 29 '11 at 23:38

The quickest way is to represent the function in terms of $1/z$ and rewrite it to a function $g(w)$ with $w=1/z$.

For example, $\frac{z^2+1}{z^2-1}=\frac{1+(1/z)^2}{1-(1/z)^2}$, so $g(w)=\frac{1+w^2}{1-w^2}=(1+w^2)(1+w^2+w^4+\cdots)=1+2w^2+2w^4+\cdots$. $w=0$ is a regular point of $g$, so $\infty$ is a regular point of $f$.

This way is justified because of $w=\frac{1}{z}$ is the holomorphic coordinate change between two charts over $S^2$, which contains $\mathbb{C}$ as an open set. Indeed, $S^2$ is a complex manifold, $\mathbb{C}$ is an open submanifold in it. We define $f$ to be holomorphic at $\infty$ if $f$ is holomophic in a neighbourhood of $\infty$ in $S^2$. This property is invariant under the coordinate change, so we can check whether $f$ is holomorphic under the coordinate $\{w\}$. This is more convenient because the coordinate $\{z\}$ has no definition at the infinity(so we just use a symbol $\infty$, this is actually saying $z$ cannot be defined at the point $\infty$ in $S^2$), while $\{w\}$ has definition at $\infty$, the coordinate of $\infty$ under the chart $\{w\}$ is $0$. Similarly, $w$ has no definition at the point $z=0$ on the sphere, or $w=\infty$ at this point.

$f$ and $g$ are just two forms of one same function under different charts $\{z\}$ and $\{w\}$. This same function can be defined to be $h(z)=\frac{z^2+1}{z^2-1}$, $h(w)=\frac{1+w^2}{1-w^2}$. This is a well-defined function on the sphere(by virtue of the relation $w=1/z$) and it's holomorphic at every point on $S^2$ except for $z=1,-1$.

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$f(z)$ is analytic at $z=\infty$ if $f(1/z)$ is analytic at $z=0$.

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Could the person who downvoted this explain? – Michael Hardy Oct 30 '11 at 23:30