f(z) = (z − 1)⁻¹ − 1 + (z − 1) − (z − 1)² + ⋯ is the series expansion of

To determine the function $f(z)$ from its given series expansion, we need to analyze the structure of the series and identify any familiar patterns, such as a geometric series.

Series Expansion Analysis

The given series expansion for $f(z)$ is:

$f(z)=(z-1{)}^{-1}-1+(z-1)-(z-1{)}^2+\dots$

This series can be rewritten by factoring out $(z-1{)}^{-1}$ from each term. This is a common technique when dealing with series that have powers of a base term.

$f(z)=(z-1{)}^{-1}\left[1-(z-1)+(z-1{)}^2-(z-1{)}^3+\dots \right]$

Upon factoring, the expression inside the square brackets clearly resembles an infinite geometric series.

Geometric Series Understanding

An infinite geometric series has the general form $a+ar+a{r}^2+a{r}^3+\dots$. The sum of such a series converges to $\frac{a}{1-r}$, provided that the absolute value of the common ratio $|r|$ is less than 1 (i.e., $|r|<1$).

Let's identify the components of the geometric series part: $1-(z-1)+(z-1{)}^2-(z-1{)}^3+\dots$

- The first term, $a$, is $1$.
- The common ratio, $r$, is $-(z-1)$. Each subsequent term is obtained by multiplying the previous term by $-(z-1)$.

Applying Geometric Series Formula

Now, we can apply the sum formula for an infinite geometric series to the bracketed part of our series, which is $S_{bracket}=\frac{a}{1-r}$:

$S_{bracket}=\frac{1}{1-(-(z-1))}$

Simplify the denominator:

$S_{bracket}=\frac{1}{1+z-1}$

$S_{bracket}=\frac{1}{z}$

**Derivation of $f(z)$ **

Finally, substitute this sum $S_{bracket}$ back into the expression for $f(z)$:

$f(z)=(z-1{)}^{-1}\times S_{bracket}$

$f(z)=\frac{1}{z-1}\times \frac{1}{z}$

$f(z)=\frac{1}{z(z-1)}$

Convergence Condition Analysis

The convergence of the geometric series part is typically valid when $|r|<1$. In this case, $|-(z-1)|<1$, which simplifies to $|z-1|<1$.

However, the options provided in the question specify the condition as $|z-1|<0$. While an absolute value can never be less than zero (it is always non-negative), we must adhere to the condition stated in the options. The function derived is $f(z)=\frac{1}{z(z-1)}$.

Let's compare our derived function and the given condition with the available options:Based on our derivation, the function is $f(z)=\frac{1}{z(z-1)}$. Option 2 matches this function form and includes the specified condition $|z-1|<0$.