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 \(a + ar + ar^2 + ar^3 + \dots\). 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:
| Option | Function and Condition |
|---|---|
| 1 | \(\frac{-1}{z(z-1)} ~for ~|z - 1| < 0\) |
| 2 | \(\frac{1}{z(z-1)}~ for~ |z - 1| < 0\) |
| 3 | \(\frac{1}{(z-1)^2} ~for ~|z - 1| < 0\) |
| 4 | \(\frac{-1}{(z-1)} ~for~ |z - 1| < 0\) |
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 \(|z - 1| < 0\).
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