Finding the Taylor Series of (1+z)/(1-z) for |z|<1

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SUMMARY

The Taylor series for the function \(\frac{1+z}{1-z}\) where \(|z| < 1\) can be derived using the geometric series formula \(\sum^{\infty}_{0} z^n = \frac{1}{1-z}\). By manipulating the series, the expression can be simplified to \(\sum^{\infty}_{0} z^n + z^{n+1}\). Further simplification reveals that this can be divided into two parts, leading to a more concise representation of the series.

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Homework Statement



Find the taylor series of [tex]\frac{1+z}{1-z}[/tex] where [tex]z[/tex] is a complex number and [tex]|z| < 1[/tex]


Homework Equations



[tex] \sum^{\infty}_{0} z^n = \frac{1}{1-z}[/tex] if [tex]|z| < 1[/tex]

The Attempt at a Solution



[tex] \sum^{\infty}_{0} z^n = \frac{1}{1-z}[/tex]

[tex]\frac{1+z}{1-z} = \sum^{\infty}_{0} z^n * (1+z) = \sum^{\infty}_{0} z^n + z^{n+1}[/tex]

I was wondering if this is as far as you can go, or if there is a more simple closed form expression for this
 
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Divide the sum into two parts and notice that

[tex]\sum_{n=0}^{\infty} z^{n+1}=\left( \sum_{n=0}^{\infty} z^{n} \right) - 1[/tex]
 

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