Write $\frac {1} {z ^ 3-iz ^ 2-z + i} $ in series of powers in a neighborhood of zero.

Question

Write $\frac {1} {z ^ 3-iz ^ 2-z + i} $ in series of powers in a neighborhood of zero.

Idea: $\frac {1} {z ^ 3-iz ^ 2-z + i} = \frac {1} {-z^2(i-z)+(i-z)}= \frac {1} {(i-z)(1-z^2)}= \frac{a}{i-z}+ \frac{bz+c}{1-z^2}$ But I don't know how to continue, can anyone help me? Please

Thanks

Answer 1

Your idea in fine. We have\begin{align}\frac1{z^3-iz^2-z+i}&=\frac12\times\left(\frac1{i-z}-\frac{z+i}{1-z^2}\right)\\&=\frac{-i}2\times\frac1{1+iz}-\frac12\times\frac{z+i}{1-z^2}\\&=-\frac i2\left(1-iz-z^2+iz^3+z^4+\cdots\right)-\frac{z+i}2\left(1+z^2+z^4+z^6+\cdots\right)\\&=-\frac i2\left(1-iz-z^2+iz^3+z^4+\cdots\right)-\frac12(i+z+iz^2+z^3+\cdots)\end{align}

Answer 2

With your approach:

$\dfrac{a}{i-z}+ \dfrac{bz+c}{1-z^2}=-ia\left(\dfrac{1}{1-(-iz)}\right)+(bz+c)\left(\dfrac{1}{1-z^2}\right) \\=\dfrac{-i}{2}\left(\dfrac{1}{1-(-iz)}\right)+(z+i)\left(\dfrac{1}{1-z^2}\right)$

Then just apply $\frac{1}{1-z^k}=\sum_{n=0}^{\infty}z^{kn}$, which works as $z$ is assumed to be in the neighborhood of zero.

Note that you can simplify furher:
$$\dfrac {1} {(i-z)(1-z^2)}=\dfrac {a} {i-z}+\dfrac{b}{1-z}+\dfrac{c}{1+z}.$$ This might although not be necessary.