I'm having a bit of difficulty figuring out this Laurent Series. The question is, find the Laurent series of $\frac{1}{(z+i)(z+1)}$ about the point $z=0$.
I have attempted to solve this by first finding the Laurent series of $\frac{1}{z+i}$ and the Laurent series of $\frac{1}{z+1}$ and then applying the termwise multiplication rule to find the series above. I have worked out the series for $\frac{1}{z+1}$ to be
$$\frac{1}{z+1}\\ \frac{1}{1+z}\\ \frac{1}{1-(-z)}\\ \sum_{n=0}^{\infty}(-z)^n\\ \sum_{n=0}^{\infty}(-1)^nz^n $$
I verified this using wolfram alpha, I know that $\frac{1}{\frac{i}{2} + 1} \approx .8-.4i$. Plugging the above formula into wolfram alpha (link) I got the same result.
I then worked out the series for $\frac{1}{z+i}$ to be $$ \frac{1}{z+i}\\ \frac{1}{i+z}\\ \frac{1}{i} \cdot \frac{1}{1+\frac{z}{i}}\\ \frac{1}{i} \cdot \frac{1}{1-iz}\\ \frac{1}{i} \cdot \sum_{n=0}^{\infty} (iz)^n\\ -i \cdot \sum_{n=0}^{\infty} i^n z^n\\ \sum_{n=0}^{\infty} (-1)^1 i^{n+1} z^n $$
Again using wolfram alpha with the point $\frac{1}{\frac{i}{2}+i} \approx -.66i$ (link) I get the same result.
So I multiply both of these series termwise by each other to obtain this series
$$ \sum_{n=0}^{\infty} (-1)^n z^n \cdot \sum_{n=0}^{\infty} (-1)^1 i^{n+1} z^n $$ $$ \sum_{n=0}^{\infty} (-1)^{n+1} \cdot i^{n+1} \cdot z^{2n} $$
However, I should get $\frac{1}{(\frac{i}{2}+1)(\frac{i}{2}+i)} \approx -.26-.53i$ but wolfram alpha gives me $.23-.94i$ instead. (link)
So, my question is where have I gone wrong? Is there a reason why I cannot termwise multiply these two series? As far as I can tell they are both series about $z=0$ so I am not sure why what I'm doing won't work.
Thanks!
In general, when you multiply two power series to get a new power series,
$$\left( \sum_{n=0}^{\infty} a_n z^n \right) \left( \sum_{n=0}^{\infty} b_n z^n \right) = \left( \sum_{n=0}^{\infty} c_n z^n \right)$$
The new series isn't a sum of term-wise product like what you have imagined. The coefficients $c_n$ of the new series is given by:
$$c_n = a_0 b_n + a_1 b_{n-1} + \cdots + a_n b_0$$
This sort of product is called Cauchy product, you can find more info on its wiki.
For the problem at hand, you should obtain the expansion by first decomposing your original expression into partial fractions:
$$\frac{1}{(z+i)(z+1)} = \frac{1}{1-i}\left(\frac{1}{z+i}-\frac{1}{z+1}\right) = \frac{1+i}{2}\left(\frac{-i}{1-iz}-\frac{1}{1+z}\right)\\ =\frac{1+i}{2}\sum_{n=0}^{\infty}\left(i^{n-1}-(-1)^n\right)z^n =\sum_{n=0}^{\infty} \left(-i z^{4n} + (i+1) z^{4n+2} - z^{4n+2}\right)$$