Question

Find the Taylor ‘s theorem expansion of 

logz=(z-1)-

(𝑧−1)

2

2

+

(𝑧−1)

3

2

− ⋯, when|𝑧 − 1| < 1.

Answer 1

We have to find Taylor's expansion of f(z) = log z

$f(z) = log(z) = log(1+z-1)$

$f(1) = log 1 = 0$

$f'(z) = \dfrac{1}{z}$ $f'(z) = \dfrac{1}{1} = 1$

$f''(z) = -\dfrac{1}{z^2}$ $f''(1) = -1$

$f'''(z) = \dfrac{2\times 1}{z^3}$ $f'''(1) = 2$

Hence, by Taylor series

$f(z) = f(a) + f'(z-a).(z-a) + \dfrac{f'''(a)(z-a)^3}{2!} +.........$

Hence,

$\implies logz = (z -1) - \dfrac{1}{2}(z -1)^2 + \dfrac{1}{3}(z -1)^3 + .........$