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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 :

This question was previously asked in
MH SET 2020 Paper 2 Philosophy Question Paper (27-Dec-2020)
The correct answer is

\(1-(z-1)+(z-1)^{2}-(z-1)^{3}+\ldots\ldots\)

Write \(z=1+(z-1)\) and use the geometric series \(\frac{1}{1+w}=\sum_{n\geq 0}(-1)^{n}w^{n}\), valid for \(|w|\lt 1\).

With \(w=z-1\): \(\frac{1}{z}=\frac{1}{1+(z-1)}=\sum_{n\geq 0}(-1)^{n}(z-1)^{n}=1-(z-1)+(z-1)^{2}-(z-1)^{3}+\cdots\), valid exactly in \(|z-1|\lt 1\), the largest disc about \(1\) avoiding the singularity at \(0\).

The series with all plus signs in powers of \(z-1\) sums to \(\frac{1}{1-(z-1)}=\frac{1}{2-z}\), a different function.

The series in powers of \(z\) are expansions about \(0\) (of \(\frac{1}{1+z}\) and \(\frac{1}{1-z}\)), not about \(1\), and \(\frac{1}{z}\) has no power series at \(0\) at all.

Check: at \(z=1.5\), \(1-0.5+0.25-0.125+0.0625-\cdots=\frac{1}{1+0.5}=0.6667=\frac{1}{1.5}\).

Hence, \(1-(z-1)+(z-1)^{2}-(z-1)^{3}+\ldots\ldots\).

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