Question

"

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

1. \(\frac{1}{z}\)
2. \(\frac{{ - 1}}{{z - 2}}\)
3. \(\frac{{z - 1}}{{z + 1}}\)
4. \(\frac{1}{{2z - 1}}\)"

Answer 1

Correct Answer - Option 1 : \(\frac{1}{z}\)

Concept:

Taylor series:

\(f(z)=f(a)+\frac{f'(a)}{1!}(z-a)+\frac{f''(a)}{2!}(z-a)^2+...\)

Sum of GP: \(f\left( z \right) = \frac{a}{{1 - r}} \)

Calculation:

f(z) = 1 + (1 – z) + (1 – z)²

The above series is in the form of G.P.

a = 1, r = (1 – z)

\(f\left( z \right) = \frac{a}{{1 - r}} = \frac{1}{{1 - \left( {1 - z} \right)}} = \frac{1}{z}\)