find-tylor-series-of-f-z-logz-about-z-o-1-i-

Question Number 148636 by tabata last updated on 29/Jul/21

f i n d t y l o r s e r i e s o f f (z) = l o g z

a b o u t z_{o} = - 1 + i

Commented by tabata last updated on 29/Jul/21

? ? ? ?

Answered by mathmax by abdo last updated on 30/Jul/21

f (z) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (z_{0})}{n!} {(z - z_{0})}^{n}

f (z) = lnz \Rightarrow f^{^{'}} (z) = \frac{1}{z} \Rightarrow f^{(n)} (z) = \frac{{(- 1)}^{n - 1} (n - 1)!}{z^{n}} \Rightarrow

f (z) = f (z_{0}) + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} (n - 1)!}{n! z_{0}^{n}} {(z - z_{0})}^{n}

= \log (- 1 + i) + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n {(- 1 + i)}^{n}} {(z + 1 - i)}^{n}

= i π + \log (e^{- \frac{i π}{4}}) - \sum_{n = 1}^{\infty} \frac{1}{n} e^{\frac{in π}{4}} {(z + 1 - i)}^{n}

logz = \frac{3 i π}{4} - \sum_{n = 1}^{\infty} \frac{1}{n} e^{\frac{in π}{4}} {(z + 1 - i)}^{n}