In the neighbourhood of z = 1, the function f(z) has a power series expansion of the form

𝑓(𝑧) = 1 + (1 − 𝑧) + (1 − 𝑧)² + ...∞

Then f(z) is

Power Series Expansion Analysis

The problem asks us to identify the function $f(z)$ based on its given power series expansion in the neighborhood of $z=1$. The expansion is provided as:

$$f(z)=1+(1-z)+(1-z{)}^2+(1-z{)}^3+\cdots$$

Our goal is to determine the closed-form expression for $f(z)$.

Geometric Series Identification

By observing the structure of the series, we can recognize it as an infinite geometric series. A geometric series has the general form $a+ar+a{r}^2+a{r}^3+\cdots$.

- **First Term ($a$):** The first term in the given series is $1$.
- **Common Ratio ($r$):** The common ratio is obtained by dividing any term by its preceding term. In this case, the ratio is $r=\frac{(1-z)}{1}=(1-z)$.

Geometric Series Sum Formula

The sum of an infinite geometric series is given by the formula:

$$S=\frac{a}{1-r}$$

This formula holds true when the absolute value of the common ratio is less than 1, i.e., $|r|<1$.

For the given series, we have $a=1$ and $r=(1-z)$. Substituting these into the sum formula:

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

Function Simplification

We can simplify the expression obtained:

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

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

Convergence Analysis

The geometric series converges if $|r|<1$. For this series, the condition is:

$$|1-z|<1$$

This inequality can be expanded as:

$$-1<1-z<1$$

Subtracting 1 from all parts gives:

$$-1-1<-z<1-1$$

$$-2<-z<0$$

Multiplying by -1 and reversing the inequalities yields:

$$0<z<2$$

This range $0<z<2$ represents a neighborhood around $z=1$. Within this neighborhood, the function $f(z)=\frac{1}{z}$ is well-defined and analytic, and its power series expansion matches the one provided in the question.

Therefore, the function is $\frac{1}{z}$.