Can anyone help with this question?
Find the power series expansions of
(a) log z about z = 1;
(b) 1/(1 + z) about z = −5.
What is the radius of convergence in each case?
Thanks for any help
For (a), where are you stuck? What have you tried?
(b) Note that $\displaystyle \frac{1}{1 + z} = \frac{1}{(z + 5) -4}$.
Region I: $\displaystyle 0 < \frac{|z + 5|}{4} < 1$.
$\displaystyle \frac{1}{(z + 5) -4} = \frac{-1}{4 - (z + 5)} = -\frac{1}{4} \left( \frac{1}{1 - \left(\frac{z+5}{4}\right)} \right)$ and you can get a series using the sum of an infinite geometric series, noting that $\displaystyle r = \frac{z + 5}{4}$.
Region II: $\displaystyle \frac{|z + 5|}{4} >1$.
$\displaystyle \frac{1}{(z + 5) -4} = \frac{1}{z + 5} \left( \frac{1}{1 - \left[ \frac{4}{z + 5}\right]} \right)$ and you can get a series using the sum of an infinite geometric series, noting that $\displaystyle r = \frac{4}{z + 5}$.