# Thread: Expansion of 1/f(z)

1. ## Expansion of 1/f(z)

Hello,
If I have an expansion of a certain function- $f(z) = \sum_{n = 0}^\infty a_n z^n$, how do I derive the expantion of 1/f(z) ?
Namely, I am looking for a series $b_m$ such that:

$\sum_{m = -\infty}^\infty b_m z^m = \frac{1}{\sum_{n = 0}^\infty a_n z^n}$

And I'm looking to express it as a function of the given series of $a_n$

Thank you

2. I don't know if this would work, but since f(z) is a complex function, you should be able to write it in terms of its real and imaginary parts, and from there be able to find $\overline{f(z)}$.

You should know that $\frac{1}{f(z)} = \frac{\overline{f(z)}}{f(z)\overline{f(z)}} = \frac{\overline{f(z)}}{\left|f(z)\right|^2}$

3. Originally Posted by dudyu
Hello,
If I have an expansion of a certain function- $f(z) = \sum_{n = 0}^\infty a_n z^n$, how do I derive the expantion of 1/f(z) ?
Namely, I am looking for a series $b_m$ such that:

$\sum_{m = -\infty}^\infty b_m z^m = \frac{1}{\sum_{n = 0}^\infty a_n z^n}$

And I'm looking to express it as a function of the given series of $a_n$

Thank you
If , then is...

(1)

... and the b coefficients are found as follows...

(2)

Kind regards

$\chi$ $\sigma$

4. Originally Posted by chisigma
If , then is...

(1)

... and the b coefficients are found as follows...

(2)

Kind regards

$\chi$ $\sigma$
Thank you!
and if $a_0$ is equal to zero, do I simply start from $a_1$ ?
Namely-
$b_n = - \frac{1}{a_1} \sum_{k = 2}^n a_k b_{n-k}$
?

5. Originally Posted by dudyu
Thank you!
and if $a_0$ is equal to zero, do I simply start from $a_1$ ?
Namely-
$b_n = - \frac{1}{a_1} \sum_{k = 2}^n a_k b_{n-k}$
?
If and You can write...

(1)

(2)

... and then compute the b coefficients imposing that...

(3)

Kind regards

$\chi$ $\sigma$