complex power series

**xifentoozlerix** · September 23rd 2008, 11:00 AM

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

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

this is my work...

$\frac{{1 + z}} {{1 - z}} = \left( {1 + z} \right)\left( {\frac{1} {{1 - z}}} \right) = \left( {1 + z} \right)\left( {\sum\limits_{n = 0}^\infty {z^n } } \right)$ $= \sum\limits_{n = 0}^\infty {z^n } + \sum\limits_{n = 0}^\infty {z^{n + 1} } = \sum\limits_{n = 0}^\infty {\left( {z^n + z^{n + 1} } \right)} = 1 + 2\sum\limits_{n = 0}^\infty {z^{n + 1} }$

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 $ \sum\limits_{n = 0}^\infty {\left( {z^n + z^{n + 1} } \right)}$ ?

**Moo** · September 23rd 2008, 11:10 AM

Hello,

Originally Posted by xifentoozlerix

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

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

this is my work...

$\frac{{1 + z}} {{1 - z}} = \left( {1 + z} \right)\left( {\frac{1} {{1 - z}}} \right) = \left( {1 + z} \right)\left( {\sum\limits_{n = 0}^\infty {z^n } } \right)$ $= \sum\limits_{n = 0}^\infty {z^n } + \sum\limits_{n = 0}^\infty {z^{n + 1} } = \sum\limits_{n = 0}^\infty {\left( {z^n + z^{n + 1} } \right)} = 1 + 2\sum\limits_{n = 0}^\infty {z^{n + 1} }$

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 $ \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 $z^{n+1}$ inside the series, but rather $z^n$ . More beautiful, since the indices are n !

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

Change the power in the second one.

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

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

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

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