Laurent series expansion of $\sqrt{\frac{z}{(z-1)(z-2)}}$ on $1<|z|<2$

Question

I think I can write the function as $$\sqrt{\frac{1}{1-1/z}}\sqrt{\frac{1}{2(1-z/2)}}$$ From here I thought about using the series expansion formula for $(1+x)^{-1/2}$, but then the product of two power series will get complicated. Is there a better way to appoach this problem?

Answer 1

Not quite a Laurent expansion, but may help:

$f(x) = {1 \over \sqrt{2}} {1 \over \sqrt{(1-{1 \over z}) (1-{z \over 2})}} = {1 \over \sqrt{2}} {1 \over \sqrt{{3 \over 2} - ({1 \over z} + {z \over 2})}} = {1 \over \sqrt{3}} {1 \over \sqrt{1 - {2 \over 3}({1 \over z} + {z \over 2}) }}$.

Let $\phi(t) = {1 \over t} + {t \over 2}$ and note that $\max_{t \in [1,2]} \phi(t) = {3 \over 2}$ and that $\phi(t) < {3 \over 2}$ for $t \in (1,2)$.

Hence if $z \in A = \{ w | 1 < |w| <2 \}$, we can use a generalised binomial expansion:

$f(z) = {1 \over \sqrt{3}} (1 - {2 \over 3}({1 \over z} + {z \over 2}) )^{-{1 \over 2}} = {1 \over \sqrt{3}} \sum_{k=0}^\infty \binom{-{1 \over 2}}{k} (- {2 \over 3})^k ({1 \over z} + {z \over 2})^k$