The given function is defined as \(f(z) = \frac{1}{1 - z - z^2}\) for \(z \in \mathbb{C}\) such that \(1 - z - z^2 \ne 0\). This function is a rational function.

Function Singularities

A rational function is analytic everywhere except at the roots of its denominator. To find the singularities (poles) of \(f(z)\), we need to find the roots of the denominator:

\(1 - z - z^2 = 0\)

We can rewrite this as \(z^2 + z - 1 = 0\). Using the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=1\), \(c=-1\):

\(z = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}\)

\(z = \frac{-1 \pm \sqrt{1 + 4}}{2}\)

\(z = \frac{-1 \pm \sqrt{5}}{2}\)

So, the singularities of \(f(z)\) are at \(z_1 = \frac{-1 + \sqrt{5}}{2}\) and \(z_2 = \frac{-1 - \sqrt{5}}{2}\). These are simple poles.

Evaluating Statements

Let's examine the given statements:

• Statement 1: "f is an entire function."

  An entire function is analytic everywhere in the complex plane. Since \(f(z)\) has singularities at \(z_1\) and \(z_2\), it is not analytic at these points. Therefore, \(f(z)\) is not an entire function. This statement is false.

• Statement 2: "f has a simple pole at z = 0."

  We found the poles are at \(z = \frac{-1 \pm \sqrt{5}}{2}\). Since \(0 \ne \frac{-1 \pm \sqrt{5}}{2}\), \(z=0\) is not a pole of \(f(z)\). In fact, the denominator \(1 - 0 - 0^2 = 1 \ne 0\), so \(f(z)\) is analytic at \(z=0\). This statement is false.

• Statements 3 and 4 concern the Taylor series expansion of \(f(z)\) around \(z=0\). Since \(z=0\) is not a singularity, \(f(z)\) is analytic at \(z=0\), and a Taylor series expansion around \(z=0\) exists. The radius of convergence of this Taylor series is the distance from \(z=0\) to the nearest singularity, which is \(|z_1| = \left|\frac{-1 + \sqrt{5}}{2}\right| = \frac{\sqrt{5}-1}{2}\).

Taylor Series Expansion

Let the Taylor series expansion of \(f(z)\) around \(z=0\) be \(f(z) = \sum_{n=0}^{\infty} a_n z^n\). We have:

\(f(z) = \frac{1}{1 - z - z^2} = \sum_{n=0}^{\infty} a_n z^n\)

Multiplying both sides by the denominator, we get:

\(1 = (1 - z - z^2) \sum_{n=0}^{\infty} a_n z^n\)

\(1 = \sum_{n=0}^{\infty} a_n z^n - z \sum_{n=0}^{\infty} a_n z^n - z^2 \sum_{n=0}^{\infty} a_n z^n\)

\(1 = \sum_{n=0}^{\infty} a_n z^n - \sum_{n=0}^{\infty} a_n z^{n+1} - \sum_{n=0}^{\infty} a_n z^{n+2}\)

Let's shift the indices in the summations to match the powers of \(z\):

\(1 = \sum_{n=0}^{\infty} a_n z^n - \sum_{n=1}^{\infty} a_{n-1} z^{n} - \sum_{n=2}^{\infty} a_{n-2} z^{n}\)

Now, let's compare the coefficients of powers of \(z\) on both sides:

• Coefficient of \(z^0\) (Constant term): \(a_0 = 1\)
• Coefficient of \(z^1\): \(a_1 - a_0 = 0 \implies a_1 = a_0\)
• Coefficient of \(z^n\) for \(n \ge 2\): \(a_n - a_{n-1} - a_{n-2} = 0 \implies a_n = a_{n-1} + a_{n-2}\) for \(n \ge 2\).

From the constant term, we have \(a_0 = 1\). From the coefficient of \(z^1\), \(a_1 = a_0 = 1\). For \(n \ge 2\), the coefficients satisfy the recurrence relation \(a_n = a_{n-1} + a_{n-2}\), which can also be written as \(a_{n+2} = a_{n+1} + a_n\) for \(n \ge 0\).

So, the Taylor series coefficients \(a_n\) are defined by the initial conditions \(a_0 = 1\), \(a_1 = 1\) and the recurrence relation \(a_{n+2} = a_{n+1} + a_n\) for \(n \ge 0\).

Comparing with Options 3 and 4

• Statement 3 says: \(a_0 = 1\), \(a_1 = 0\) and \(a_{n+2} = a_n + a_{n+1}\) for \(n \ge 0\). This is incorrect because \(a_1\) should be 1, not 0.
• Statement 4 says: \(a_0 = 1\), \(a_1 = 1\) and \(a_{n+2} = a_{n} + a_{n+1}\) for \(n \ge 0\). This matches our derived initial conditions and recurrence relation.

Therefore, statement 4 is the true statement.