Exercise
Suppose that
$ f(z) = \frac{2}{z+2i} - \frac{5}{z-4} $
Find the Laurent series for f in powers of (z - 2) that converges when z = i.
Given solution
(1) $ f(z) = \frac{2}{z+2i} - \frac{5}{z-4} $
(2) $ f(z) = \frac{2}{(z-2)(2+2i)} - \frac{5}{(z-2)-2} $
(3) $ f(z) = \frac{2}{2+2i}\frac{1}{1+\frac{z-2}{2+2i}} - \frac{5}{(z-2)}\frac{1}{(1-\frac{2}{z-2})} $
(4) $ f(z) = \frac{2}{2+2i}\frac{1}{1+\frac{z-2}{2+2i}} - \frac{5}{(z-2)}\frac{1}{(1-\frac{2}{z-2})} $
(5) $ f(z) = \frac{2}{2+2i}(1 - \frac{z-2}{2+2i} + (\frac{z-2}{2+2i})^2 - (\frac{z-2}{2+2i})^3 + ...) - \frac{5}{z-2}(1 + \frac{2}{z-2} + (\frac{2}{z-2})^2 + (\frac{2}{z-2})^2 + ...)$
I suspect that there may be an error in the text where the 2 power repeats which it should be 3.
(6) $ \sum\limits_{k \in Z} c_k(z-2)^k \text{, where } c_k=\frac{2(-1)^{-k}}{(2+2i)^{k+1}} \text{where } k \ge 0 \text{ and } c_k = \frac{-5}{2^{k+1}} \text{ where k } \lt 0$
I'm not very familiar with Laurent series beyond the basic theory of what they are but I can follow steps 1-4, however, I'm struggling to follow the step 4->5.
Why does $ \frac{1}{1+\frac{z-2}{2+2i}} = (1 - \frac{z-2}{2+2i} + (\frac{z-2}{2+2i})^2 - (\frac{z-2}{2+2i})^3 + ...)$ and $ \frac{1}{(1-\frac{2}{z-2})} = (1 + \frac{2}{z-2} + (\frac{2}{z-2})^2 + (\frac{2}{z-2})^2 + ...)$?
The solution doesn't seem to explain the step and I can't intuitively see why they equal those sequences.
I had a look here (wolfram series expansion) and saw the expansion for $ \frac{1}{1 - x}$ but the signs don't alternate like in the solution above. I suspect that this is related to $i$ being in the denominator as the second component doesn't have alternating signs but I'm not very familiar with sequences and series.
First thing is you are using geometric series $\frac{1}{1-z}=1+z+z^2+...$. This is true only for $\mid z\mid < 1$. At $z=\iota$, $\mid \frac{2}{\iota-2}\mid = \frac{2}{\sqrt{5}} <1$ and $\mid \frac{\iota-2}{2+2\iota}\mid = \frac{\sqrt{5}}{\sqrt{8}} <1$. So thats why we can use the geometric series formula for those two above.
Regarding your second problem, you should first write $\frac{1}{1+\frac{z-2}{2+2\iota}}= \frac{1}{1-(-\frac{z-2}{2+2\iota})}$ and then expand it using geometric series. This is because geometric formula is for function $\frac{1}{1-z}$ and not for type $\frac{1}{1+z}$. That is why the alternate signs are coming.