Looking for the Laurent series of
$f(z)=\frac{1}{z(z-1)(3i-z)}=\frac{i}{3 z} + \frac{\frac{1}{10} - \frac{i}{30}}{z - 3 i} - \frac{\frac{1}{10} + \frac{3i}{10}}{z - 1}$
So we look for $z_0=0,z_1=1,z_2=3i$ on
$A=\{z \in \Bbb C: |z| <1\}$
$B=\{z \in \Bbb C: 1< |z| <3\}$
$C=\{z \in \Bbb C: 3< |z| < \infty\}$
On $A$:
$\frac{i}{3 z}=\frac{i}{3 (z-1+1)}=\frac{i}{3 }\sum_{n=0}^\infty \frac{i}{3 }(z-1)^n(-1)^n $
$\frac{\frac{1}{10} - \frac{i}{30}}{z - 3 i} =\frac{1}{z}\frac{\frac{1}{10} - \frac{i}{30}}{1 - 3 i/z}=\frac{1}{z }\sum_{n=0}^\infty (\frac{1}{10} - \frac{i}{30})(-\frac{3i}{z })^n=\sum_{n=0}^\infty (\frac{1}{10} - \frac{i}{30})((-\frac{3i}{z })^{n+1}\sum_{n=1}^\infty (\frac{1}{10} - \frac{i}{30})((-\frac{3i}{z })^n$
$\frac{\frac{1+3i}{10}}{z - 1}=-\frac{1+3i}{10}\sum_{n=0}^\infty z^n $
Is this correct and how is this different to B and C ? How would I begin with in B ?
Hint: for $|z|<1$ we have $$\frac{1}{1-z} = \sum_{n=0}^\infty z^n$$ but for $|z|>1$ we have $$\frac{1}{1-z} =-\frac{1}{z} \frac{1}{1-\frac{1}{z}} = -\sum_{n=1}^\infty\frac{1}{z^n}$$