1. ## complex power series

the problem is to find the power series about the origin of the given function:

$\displaystyle \frac{{1 + z}} {{1 - z}},{\text{ }}\left| z \right| < 1$

this is my work...

$\displaystyle \frac{{1 + z}} {{1 - z}} = \left( {1 + z} \right)\left( {\frac{1} {{1 - z}}} \right) = \left( {1 + z} \right)\left( {\sum\limits_{n = 0}^\infty {z^n } } \right) $$\displaystyle = \sum\limits_{n = 0}^\infty {z^n } + \sum\limits_{n = 0}^\infty {z^{n + 1} } = \sum\limits_{n = 0}^\infty {\left( {z^n + z^{n + 1} } \right)} = 1 + 2\sum\limits_{n = 0}^\infty {z^{n + 1} } my question is if this is the form required or is there is some way to get the 1 back inside the sum? or would it be better to leave it as \displaystyle \sum\limits_{n = 0}^\infty {\left( {z^n + z^{n + 1} } \right)} ? 2. Hello, Originally Posted by xifentoozlerix the problem is to find the power series about the origin of the given function: \displaystyle \frac{{1 + z}} {{1 - z}},{\text{ }}\left| z \right| < 1 this is my work... \displaystyle \frac{{1 + z}} {{1 - z}} = \left( {1 + z} \right)\left( {\frac{1} {{1 - z}}} \right) = \left( {1 + z} \right)\left( {\sum\limits_{n = 0}^\infty {z^n } } \right)$$\displaystyle = \sum\limits_{n = 0}^\infty {z^n } + \sum\limits_{n = 0}^\infty {z^{n + 1} } = \sum\limits_{n = 0}^\infty {\left( {z^n + z^{n + 1} } \right)} = 1 + 2\sum\limits_{n = 0}^\infty {z^{n + 1} }$

my question is if this is the form required or is there is some way to get the 1 back inside the sum? or would it be better to leave it as $\displaystyle \sum\limits_{n = 0}^\infty {\left( {z^n + z^{n + 1} } \right)}$?
It's only my opinion. Maybe your teacher views it a different way, maybe you prefer it a certain way...

It's better not to leave $\displaystyle z^{n+1}$ inside the series, but rather $\displaystyle z^n$. More beautiful, since the indices are n !

$\displaystyle =\sum_{n=0}^\infty z^n+\sum_{n=0}^\infty z^{n+1}$

Change the power in the second one.

$\displaystyle =\sum_{n=0}^\infty z^n+\sum_{n={\color{red}1}}^\infty z^n$

$\displaystyle =1+2 \sum_{n={\color{red}1}}^\infty z^n$

or $\displaystyle =\left(2 \sum_{n=0} z^n \right)-1$