Laurent series of function, partial fraction decomp. problem

Question

Find the Laurent series of $f(z)$ in the region $1<|z-i|<\sqrt{2}$ $$f(z) = \frac{1}{z(z-i)^2(z-1)} $$ The region doesn't include any of the singularities hence expansion is possible
The problem is there is too much going on, I don't know what to do with the $\frac{1}{(z-i)^2}$ term in this situation. I could look at two separate factors $$\frac{1}{z(z-1)}\cdot \frac{1}{(z-i)^2} = \frac{A}{z(z-1)} + \frac{B}{(z-i)^2}, A,B\in\mathbb{C}$$ but I can't determine $A,B$. Assuming $A,B$ are determined the first series is easy and for the second we can use the fact that $-\left (\frac{1}{z-i}\right )' = \frac{1}{(z-i)^2}$.

Trouble is, what to do with $A,B$? And how would the final series look like? Can we just add the two series together?

Alternative ideas also welcome.

Attempt:
We have $$A(z-i)^2 + Bz(z-1) = 1 = Az^2 -2Aiz -A + Bz^2 -Bz $$ We know that $(2Ai+B)z = 0\Longrightarrow 2Ai + B=0$, meaning $B=-2Ai$,
but now I don't see where I would get enough information to determine what $A$ is.

Scratch that - unnecessary work

Answer 1

The Laurent expansion around $z=i$ will be in powers of $z-i$. You have to do nothing with the factor $\dfrac{1}{(z-i)^2}$, since it is already a power of $z-i$. $$ \frac{1}{z(z-1)}\cdot \frac{1}{(z-i)^2} = \frac{1}{(z-i)^2}\Bigl(\frac{A}{z}+\frac{B}{z-1}\Bigr)\quad A,B\in\mathbb{C}. $$

Answer 2

It is easy to calculate the Laurent series using the equation that expresses the coefficients in terms of a line integral and the residue theorem. We have

$$ a_n = \frac{1}{2\pi i} \oint_{\gamma} \frac{f(z)}{(z-i)^{n+1}} \, dz = \frac{1}{2\pi i} \oint_{\gamma} \frac{1}{(z-i)^2 z (z - 1) (z - i)^{n+1}} \, dz \\ = \frac{1}{2\pi i} \oint_{\gamma} \frac{1}{(z - i)^{n + 3} z (z - 1)} \, dz $$

where $\gamma$ is a simple closed curve that lies in the annulus $1 < |z - i| < \sqrt{2}$ and winds around $z = i$ once counter clockwise. It also winds around $z = 0$ and so by the residue theorem we have

$$ a_n = \operatorname{Res} \left( \frac{1}{(z - i)^{n + 3} z (z - 1)}, i \right) + \operatorname{Res} \left( \frac{1}{(z - i)^{n + 3} z (z - 1)}, 0 \right) \\ = \frac{\delta_{n,-4}}{i(i - 1)} + \frac{1}{(-i)^{n+3}(-1)} = -(i)^{n+3} + \delta_{n,-4} \frac{-1 + i}{2}$$

and so

$$ f(z) = \frac{-1 + i}{2} \frac{1}{(z - i)^4} - \sum_{n=-\infty}^{\infty} \frac{i^{n+3}}{(z-i)^n}. $$

Answer 3

I hope I haven't messed up the signs somewhere.
We start with: $$f(z) =\frac{1}{(z-i)^2}\left (\frac{A}{z} +\frac{B}{z-1}\right ) = \frac{1}{(z-i)^2}\left (-\frac{1}{z} +\frac{1}{z-1}\right ) $$ we proceed by finding the Laurent series for $-1/z$ $$-\frac{1}{z} = \frac{1}{-z+i-i} = \frac{1}{-(z-i)-i} = -\frac{1}{z-i}\cdot\frac{1}{1-\left (-\frac{i}{z-i}\right )} =-\frac{1}{z-i}\sum_{n=0}^\infty (-1)^n\frac{i^n}{(z-i)^n} =\sum_{n=0}^\infty (-1)^{n+1}\frac{i^n}{(z-i)^{n+1}} $$ which converges when $|z-i|>1$
Then, as the singularity of $1/(z-1)$ is at $z=1$, we need a Taylor series $$\frac{1}{z-1} =-\frac{1}{(1-i)-(z-i)} =-\frac{1}{1-i}\cdot\frac{1}{1 -\frac{z-i}{1-i}}=-\frac{1}{1-i}\sum_{n=0}^\infty \frac{(z-i)^n}{(1-i)^n} =-\sum_{n=0}^\infty \frac{(z-i)^n}{(1-i)^{n+1}} $$ which converges when $\lvert\frac{z-i}{1-i}\rvert< 1\Longrightarrow |z-i|<|1-i|=\sqrt{2}$.
Finally, putting the two things together $$f(z) = \sum_{n=0}^\infty (-1)^{n+1}\frac{i^n}{(z-i)^{n+3}} - \sum_{n=0}^\infty \frac{(z-i)^{n-2}}{(1-i)^{n+1}},\ \ \ 1<|z-i|<\sqrt{2} $$

Do I have to combine the sums into one big sum or would a final answer in such a form be acceptable, provided I have done everything correctly?