Answer to Question #263476 in Complex Analysis for Rajappriya

Question

Develop 1/(1+z^2) in powers of z-a, a being a real number.Find the general coefficient and for a=1 reduce to simplest form.

Answer 1

Taylor series:

$f(z)=\displaystyle{\sum_{k=0}^{\infin}}\frac{f^{(k)}(a)}{k!}(z-a)^k$

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

$f''(z)=\frac{6z^2-2}{(1+z^2)^3}$

$f'''(z)=\frac{24z(1-z^2)}{(1+z^2)^4}$

$\frac{1}{1+z^2}=\frac{1}{1+a^2}-\frac{2a}{(1+a^2)^2}(z-a)+\frac{6a^2-2}{2(1+a^2)^3}(z-a)^2+\frac{24a(1-a^2)}{6(1+a^2)^4}(z-a)^3+...$

for a=1:

$\frac{1}{1+z^2}=1-\frac{z}{2}+\frac{(z-1)^2}{4}-\frac{(z-1)^4}{8}+\frac{(z-1)^5}{8}-$