Question

= t a r has a Laurent series representation about the 2. Show that f(s) = point = = i given by ř (1 - i)n-1 f(3) = (valid for 12-il > V2).

Answer 1

Solution:

Given that

$f(z)=\frac{1}{1-z_{0}}$$[\because \frac{1}{1-z_{0}}=\frac{1}{1-z}]$

$f(z)=\frac{1}{1-i}*\frac{1}{1-\frac{z-i}{1-i}}$

Here, it has the laurent series representation about point i.e, $z_{0}=i$

Now,

$f(z)=\frac{1}{1-i}*\frac{1}{1-\frac{z-i}{1-i}}$

  $\left | z-i \right |>\sqrt{2}>1$

$\therefore \left | \frac{1}{z-i} \right |<1]$

$\therefore$ The series

$f(z)=-\frac{1}{z-i}*\frac{1}{1-\frac{1-i}{z-i}}$

$f(z)=-\frac{1}{z-i}*[1-(\frac{1-i}{z-i})]^{-1}$

$f(z)=-\frac{1}{z-i}*[1+(\frac{1-i}{z-i})+(\frac{1-i}{z-i})^{2}+........]$

$f(z)=-\frac{1}{z-i}\sum_{n=1}^{4}\left ( \frac{1-i}{z-i} \right )^{n-1}$

$\therefore \left [ \left | \frac{1}{z-i} \right |<1 \right ]$

$f(z)=-\sum_{n=1}^{4}\frac{(1-i)^{n-1}}{(z-i)^{n}}$

Hence,

we proved that

$f(z)=\frac{1}{1-z_{0}}=-\sum_{n=1}^{4}\frac{(1-i)^{n-1}}{(z-i)^{n}}$

(valid for $\left | z-i \right |>\sqrt{2}$ )