The question asks about the properties of the difference of two entire functions, \(f\) and \(g\), given a specific condition on their limit as \(z\) approaches infinity.

We are given that \(f\) and \(g\) are entire functions, and for some fixed positive integer \(n\), we have:

• \(\lim _{z \rightarrow \infty} \frac{f(z)}{z^n}=1\)
• \(\lim _{z \rightarrow \infty} \frac{g(z)}{z^n}=1\)

An entire function is analytic everywhere in the complex plane. The behavior of an entire function as \(z \rightarrow \infty\) tells us a lot about the function itself. Specifically, if an entire function grows no faster than a polynomial of degree \(k\), it must be a polynomial of degree at most \(k\).

Entire Functions and Limit at Infinity

Let's analyze the condition \(\lim _{z \rightarrow \infty} \frac{f(z)}{z^n}=1\). This means that for large values of \(|z|\), \(f(z)\) behaves very much like \(z^n\). More formally, we can say that \(f(z) = z^n + h(z)\), where \(h(z)\) is an entire function such that \(\lim _{z \rightarrow \infty} \frac{h(z)}{z^n}=0\).

Consider an entire function \(F(z)\). Its behavior at infinity can be analyzed by considering the function \(G(w) = F(1/w)\) around \(w=0\). If \(F(z)\) is entire, \(G(w)\) is analytic for \(w \neq 0\). The behavior of \(F(z)\) at \(z = \infty\) corresponds to the behavior of \(G(w)\) at \(w = 0\).

An entire function \(F(z)\) has a Laurent series expansion around infinity of the form \(F(z) = \sum_{k=0}^\infty a_k z^k\). The condition \(\lim_{z \rightarrow \infty} \frac{F(z)}{z^n} = C\) for some constant \(C\) implies constraints on the coefficients \(a_k\).

Given \(\lim_{z \rightarrow \infty} \frac{f(z)}{z^n}=1\), we have \(\lim_{z \rightarrow \infty} \frac{\sum_{k=0}^\infty a_k z^k}{z^n} = \lim_{z \rightarrow \infty} \sum_{k=0}^\infty a_k z^{k-n} = 1\).

For this limit to exist and be equal to 1, several things must be true:

• For \(k > n\), the term \(a_k z^{k-n}\) would go to infinity as \(z \rightarrow \infty\) unless \(a_k = 0\). Thus, \(a_k = 0\) for all \(k > n\). This means the series terminates, and \(f(z)\) must be a polynomial of degree at most \(n\).
• For \(k < n\), the term \(a_k z^{k-n}\) would go to 0 as \(z \rightarrow \infty\). These terms do not affect the limit's constant value.
• For \(k = n\), the term \(a_n z^{n-n} = a_n z^0 = a_n\). This is the only term that can contribute a non-zero constant to the limit.

So, for the limit to be 1, we must have \(a_k = 0\) for \(k > n\), and \(a_n = 1\). The function \(f(z)\) must therefore be a polynomial of degree exactly \(n\), with the coefficient of \(z^n\) being 1.

Let \(f(z) = z^n + a_{n-1} z^{n-1} + \dots + a_1 z + a_0\).

Similarly, since \(\lim _{z \rightarrow \infty} \frac{g(z)}{z^n}=1\), \(g(z)\) must also be a polynomial of degree exactly \(n\), with the coefficient of \(z^n\) being 1.

Let \(g(z) = z^n + b_{n-1} z^{n-1} + \dots + b_1 z + b_0\).

Function Difference: f - g

Now let's consider the difference \(f(z) - g(z)\):

\(f(z) - g(z) = (z^n + a_{n-1} z^{n-1} + \dots + a_0) - (z^n + b_{n-1} z^{n-1} + \dots + b_0)\)

\(f(z) - g(z) = (a_{n-1} - b_{n-1}) z^{n-1} + (a_{n-2} - b_{n-2}) z^{n-2} + \dots + (a_0 - b_0)\)

This is a polynomial. The highest possible power of \(z\) is \(n-1\), which occurs if \(a_{n-1} \neq b_{n-1}\). If \(a_{n-1} = b_{n-1}\), the coefficient of \(z^{n-1}\) is zero, and the degree is less than \(n-1\). In any case, the degree of the polynomial \(f(z) - g(z)\) cannot exceed \(n-1\).

Thus, \(f - g\) is necessarily a polynomial of degree at most \(n-1\).

Evaluating the Options

Let's check this against the given options:

1. f = g: This is not necessarily true. For example, take \(n=2\), \(f(z) = z^2 + z\) and \(g(z) = z^2\). Both satisfy the condition, but \(f \neq g\).
2. f - g is necessarily a polynomial of degree at most n - 1: This matches our conclusion.
3. there exist f, g with these properties such that f - g is a polynomial of degree n: This is false, as the highest possible degree of \(f-g\) is \(n-1\).
4. there exist f, g with these properties such that f - g is not a polynomial: This is false, as both \(f\) and \(g\) are polynomials, and their difference is always a polynomial.

Therefore, the statement that \(f - g\) is necessarily a polynomial of degree at most \(n - 1\) is the true statement.