Analytic Functions at Infinity

The question asks to identify the true statement regarding functions that are analytic at infinity ($\infty$). The definition provided states that a function \(f : \mathbb{C} \to \mathbb{C}\) is analytic at $\infty$ if the function \(g(w) = f\left(\frac{1}{w}\right)\) is analytic at \(w=0\).

Being analytic at a point \(w_0\) means that the function can be represented by a power series in a neighborhood of \(w_0\). For \(g(w)\) to be analytic at \(w=0\), it must be defined and have a convergent power series expansion in some open disk centered at 0, i.e., for \(|w| < \delta\) for some \(\delta > 0\).

Analyzing Each Statement

Statement 1: Any non-constant polynomial is analytic at ∞

Let \(f(z) = a_n z^n + a_{n-1} z^{n-1} + \dots + a_1 z + a_0\) be a non-constant polynomial, where \(a_n \ne 0\) and \(n \ge 1\).

The function \(g(w)\) is given by \(g(w) = f\left(\frac{1}{w}\right)\):

\[g(w) = a_n \left(\frac{1}{w}\right)^n + a_{n-1} \left(\frac{1}{w}\right)^{n-1} + \dots + a_1 \left(\frac{1}{w}\right) + a_0\]

\[g(w) = \frac{a_n}{w^n} + \frac{a_{n-1}}{w^{n-1}} + \dots + \frac{a_1}{w} + a_0\]

For a non-constant polynomial (\(n \ge 1, a_n \ne 0\)), the term \(\frac{a_n}{w^n}\) has a pole of order \(n\) at \(w=0\). A function with a pole at a point is not analytic at that point. Therefore, \(g(w)\) is not analytic at \(w=0\).

So, any non-constant polynomial is not analytic at ∞. Statement 1 is false.

Statement 2: If f is analytic at ∞ then f is bounded

If \(f\) is analytic at ∞, then \(g(w) = f(1/w)\) is analytic at \(w=0\). If \(g(w)\) is analytic at \(w=0\), it is defined and analytic in some open disk \(D(0, \delta)\) for \(\delta > 0\). A function analytic on a disk is bounded on any closed disk contained within that disk. Thus, \(g(w)\) is bounded for \(|w| \le \delta/2\). This means \(|f(1/w)| \le M\) for some constant \(M\) whenever \(|w| \le \delta/2\).

Let \(z = 1/w\). The condition \(|w| \le \delta/2\) is equivalent to \(|1/z| \le \delta/2\), which means \(|z| \ge 2/\delta\). So, \(f(z)\) is bounded for all \(z\) outside a certain closed disk centered at the origin (i.e., for large values of \(|z|\)).

However, the statement says "f is bounded". This could mean bounded on its entire domain. Consider the function \(f(z) = 1/z\). This function is analytic on \(\mathbb{C} \setminus \{0\}\). Let's check if it's analytic at ∞.

\[g(w) = f\left(\frac{1}{w}\right) = \frac{1}{1/w} = w\]

The function \(g(w)=w\) is analytic at \(w=0\). Thus, \(f(z) = 1/z\) is analytic at ∞. But \(f(z)=1/z\) is not bounded on its domain \(\mathbb{C} \setminus \{0\}\) because as \(z \to 0\), \(|f(z)| \to \infty\).

So, if "f is bounded" means bounded on its entire domain, Statement 2 is false. If it means bounded for large \(|z|\), it is true. Given the other options and the likely context, "bounded" probably refers to the entire domain. Statement 2 is likely false.

Statement 3: For any \(z_0\) in \(\mathbb{C}\), the function \(f(z)=e^{\frac{1}{z-z_0}}\) is analytic at ∞

We need to check if \(g(w) = f(1/w)\) is analytic at \(w=0\). Let \(f(z) = e^{\frac{1}{z-z_0}}\).

\[g(w) = f\left(\frac{1}{w}\right) = e^{\frac{1}{\frac{1}{w} - z_0}}\] \[g(w) = e^{\frac{1}{\frac{1-z_0w}{w}}} = e^{\frac{w}{1-z_0w}}\]

The function \(g(w) = e^{\phi(w)}\), where \(\phi(w) = \frac{w}{1-z_0w}\).

The exponential function \(e^u\) is analytic everywhere in \(\mathbb{C}\).

The function \(\phi(w) = \frac{w}{1-z_0w}\) is a rational function. A rational function is analytic everywhere its denominator is non-zero.

Case 1: \(z_0 = 0\). \(f(z) = e^{\frac{1}{z}}\). \(g(w) = e^{\frac{w}{1-0 \cdot w}} = e^w\). The function \(g(w) = e^w\) is analytic everywhere in \(\mathbb{C}\), including at \(w=0\).

Case 2: \(z_0 \ne 0\). The denominator of \(\phi(w)\) is \(1-z_0w\). This is zero when \(1-z_0w = 0\), which means \(w = 1/z_0\). Since \(z_0 \ne 0\), \(1/z_0\) is a finite non-zero value. The function \(\phi(w) = \frac{w}{1-z_0w}\) is analytic at \(w=0\) because the denominator \(1-z_0 \cdot 0 = 1 \ne 0\). Since \(\phi(w)\) is analytic at \(w=0\), and the exponential function is analytic, the composition \(g(w) = e^{\phi(w)}\) is also analytic at \(w=0\).

In both cases (\(z_0 = 0\) and \(z_0 \ne 0\)), \(g(w) = f(1/w)\) is analytic at \(w=0\). Therefore, \(f(z) = e^{\frac{1}{z-z_0}}\) is analytic at ∞ for any \(z_0 \in \mathbb{C}\).

Statement 3 is true.

Statement 4: Any entire function can be extended to an analytic function at ∞

An entire function is a function that is analytic on the entire complex plane \(\mathbb{C}\). For an entire function \(f(z)\) to be analytic at ∞, \(g(w) = f(1/w)\) must be analytic at \(w=0\).

Consider the entire function \(f(z) = z\). \(g(w) = f\left(\frac{1}{w}\right) = \frac{1}{w}\). The function \(g(w) = 1/w\) has a pole at \(w=0\). It is not analytic at \(w=0\). So, \(f(z)=z\) is an entire function that is not analytic at ∞.

Consider the entire function \(f(z) = e^z\). \(g(w) = f\left(\frac{1}{w}\right) = e^{\frac{1}{w}}\). The function \(g(w) = e^{1/w}\) has an essential singularity at \(w=0\). It is not analytic at \(w=0\).

In fact, the only entire functions that are analytic at ∞ are the constant functions. This is because if \(f(z)\) is entire and analytic at ∞, it is analytic on the extended complex plane \(\mathbb{C} \cup \{\infty\}\). Such a function is necessarily constant by Liouville's theorem extended to the sphere.

So, not any entire function can be extended to an analytic function at ∞. Statement 4 is false.

Conclusion

Based on the analysis of each statement, only Statement 3 is true.

The function \(f(z) = e^{\frac{1}{z-z_0}}\) for any \(z_0 \in \mathbb{C}\) is analytic at ∞.