The problem asks for the radius of convergence of the Taylor series of a given rational function \(f(z)\) centered at \(z=1\).

The function is given by: \[f\left( z \right) = \frac{{{z^3} + 2z - 4}}{z}\] This is a rational function because it is a ratio of two polynomials, \(z^3 + 2z - 4\) and \(z\).

Rational Function Singularities

A rational function is analytic everywhere in the complex plane except at the points where its denominator is zero. These points are called singularities of the function.

For the given function \(f(z)\), the denominator is \(z\). The denominator is zero when \(z = 0\).

Thus, the function \(f(z)\) has only one singularity at \(z = 0\). The function is analytic everywhere else in the complex plane (\(\mathbb{C} \setminus \{0\}\)).

Taylor Series and Radius of Convergence

The Taylor series of a function \(f(z)\) centered at a point \(z_0\) converges in the largest open disk centered at \(z_0\) where the function \(f(z)\) is analytic.

The radius of convergence of the Taylor series is the distance from the center of the expansion \(z_0\) to the nearest point where the function is not analytic (a singularity).

Radius Calculation

In this problem, the Taylor series is centered at \(z_0 = 1\).

The function \(f(z)\) has a singularity at \(z = 0\).

The distance between the center of the expansion \(z_0 = 1\) and the singularity at \(z = 0\) is given by the modulus of the difference: \[\left|1 - 0\right| = \left|1\right| = 1\]

Since \(z=0\) is the only singularity, the distance to the nearest singularity from the center \(z=1\) is 1.

Therefore, the radius of convergence of the Taylor series of \(f(z)\) at \(z = 1\) is 1.

This means the Taylor series converges for all \(z\) such that \(|z - 1| < 1\).