<s>[INST] Find Laurent series expansion of $\sqrt{(z-1)(z-2)}, |z|>2$ 
Find Laurent series expansion of $\sqrt{(z-1)(z-2)}$, $|z|>2$.

Using the The Binomial Theorem: $\pm z(z-1)^{1/2}(z-2)^{1/2} = \pm z[\sum_{k=0}^\infty\binom{1/2}{k}(-1)^{k}z^k][\sum_{k=0}^\infty\binom{1/2}{k}(-2)^{k}z^k]$.
I have a problem how multiply these two sums. 
The answer in the book: $c_n$ = $\binom{1/2}{k}$
+ $2\binom{1/2}{k-1}\binom{1/2}{1}$ + $2^2\binom{1/2}{k-2}\binom{1/2}{2}$ + … + $2^n\binom{1/2}{k}$ and  expansion is $\pm [c_0z-c_1+\frac{c_2}{z}-\frac{c_3}{z^2}+...]$
Could someone, please, help to get the right answer? 
I think Cauchy product of the series can help to find coeffitients.
 [/INST]    
The idea was correct:
$z(z-1)^{1/2}(z-2)^{1/2} =\pm z\sqrt{(1-\frac{1}{z})}\sqrt{(1-\frac{2}{z})}$, 
then using the The Binomial Theorem: 
$\sqrt{(1-\frac{1}{z})} = [\sum_{k=0}^\infty\binom{1/2}{k}(-1)^{k}z^{-k}] = \sum_{k=0}^\infty(-1)^{k}\binom{1/2}{k}z^{-k}$ and
$\sqrt{(1-\frac{2}{z})} = [\sum_{k=0}^\infty\binom{1/2}{k}(-2)^{k}z^{-k}] = \sum_{k=0}^\infty(-1)^{k}2^k\binom{1/2}{k}z^{-k}$
Now we have : $\pm z[\sum_{k=0}^\infty(-1)^{k}\binom{1/2}{k}z^{-k}][\sum_{k=0}^\infty(-1)^{k}2^k\binom{1/2}{k}z^{-k}]$
Using Cauchy product: $[\sum_{k=0}^\infty a_kx^n * \sum_{k=0}^\infty b_kx^n] = \sum_{k=0}^\infty c_kx^n$,
and $c_0 = a_0b_0, c_1 = a_1b_0 + a_0b_1  $ etc
$c_n = \sum_{k=0}^n a_{n-k}b_k $.
Then the coeff. $c_n  =  \sum_{k=0}^n(-1)^{n-k}\binom{1/2}{n-k}(-1)^k2^k\binom{1/2}{k} = (-1)^n\sum_{k=0}^n2^{k}\binom{1/2}{n-k}\binom{1/2}{k}$.
The Laurent  expansion: $\pm[c_0z - c_1 + \frac{c_2}{z^2} - ... ]$
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