How can we find the Laurent Series of $\sqrt{(z-1)(z-2)}$ near point $z=\infty$? By definition, it's equivalent to asking the Laurent Seires of $\sqrt{\frac{(t-1)(t-2)}{t^2}}$ near the point $t=0$. The singularities are $t=0,1, 2$ but the square-root function is multivalued. So we have to define a branch cut $[-\infty,0]$, $[1,2]$. But all points on the branch cut is obviously non-analytic. So can we say the Laurent Series do not exist?
The Laurent Series of $\sqrt{(z-1)(z-2)}$ at the point $z=\infty$?
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complex-analysis
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0@KCd I see, you mean it's not possible to define the Laurent Series at points like $z=0$ for $\sqrt{z}$? – 2012-05-20
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You can define a branch of $\sqrt{(z-1)(z-2)}$ that has its branch cut on the interval $[1,2]$, and is thus analytic in a neighbourhood of $\infty$. If $z = 1/t$, we write $(z-1)(z-2) = (1-t)(1-2t)/t^2$ and take $\sqrt{(z-1)(z-2)} = \sqrt{(1-t)(1-2t)}/t$. Note that $\sqrt{(1-t)(1-2t)}$ (using the principal branch of the square root) is analytic in a neighbourhood of $t=0$.
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0Right, $\sqrt{(t-1)(t-2)/t}$ has a branch point at $0$ because $\sqrt{1/t}$ does. – 2012-05-20