The Laurent series around $z=0$ of the function $f(z) = \frac{z}{(z-i)(z-2)}$ in the annulus $A(0,1,2)$

Question

What I got so far: $$ \frac{z}{(z-i)(z-2)} = \frac{z}{(2-i)(z-i)} + \frac{z}{(2-i)(z-2)} $$

which is equal to

$$ \frac{z}{(2-i)(z-i)} + \frac{z}{(2-i)(z-2)} = \frac{z}{(2-i)z + 1-2i} + \frac{z}{(2-i)z + 2i - 4} $$ and

$$ \frac{z}{(2-i)z + 1-2i} + \frac{z}{(2-i)z + 2i - 4} = \frac{1}{z(2-i)}\bigg(\frac{1}{1 + \frac{1 - 2i}{z(2-i)}} - \frac{1}{1 + \frac{2i - 4}{z(2-i)}}\bigg) $$

Now if we just look at

$$ \frac{1}{1 + \frac{2i - 4}{z(2-i)}} = \frac{1}{1 - \frac{2 - i}{z}} $$

and this reminds me of a geometric progression. But I don't know how to continu.. Could anyone finish this for me or give me tips? Thanks.

Answer 1

(z/((z-i)(z-2)))=(z/((2-i)(z-i)))+(z/((2-i)(z-2))),1<∣z∣<2. The expansion in powers of z is 
(z/((2-i)z(1-i/z)))-(z/((2-i)2(1-z/2)))

=(1/(2-i))∑₀((i/z))ⁿ-(1/(4-2i))∑₀((z/2))ⁿ.