Discussion
Define \(f : \mathbf{C} - \{0\} \to \mathbf{C}\) by \(f(z) = \dfrac{1}{z}\). Then the power series expansion of \(f(z)\) about \(z = 1\) in the disc \(\{z : |z - 1| \lt 1\}\) is :
Answer: A
The expansion is \(1 - (z - 1) + (z - 1)^{2} - (z - 1)^{3} + \cdots\).
Write \(z = 1 + (z - 1)\) and use the geometric series, valid when \(|z - 1| \lt 1\):
\[ \begin{aligned} \frac{1}{z} &= \frac{1}{1 + (z - 1)} \\ &= \sum_{n \ge 0} (-1)^{n}(z - 1)^{n} \end{aligned} \]
The series with all plus signs would be \(1/(2 - z)\), and the two options in powers of \(z\) are expansions about 0 of \(1/(1 + z)\) and \(1/(1 - z)\), not of \(1/z\) about 1.
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