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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?

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You don't have to choose the specific branch cuts you made; it is not mandatory to make the choice you did. And I don't think $\infty$ is really the primary problem for you. Let's take a simpler situation. Based on the concerns you have, what would you say is the Laurent series for $\sqrt{z}$ at $z = 1$? –  KCd May 20 '12 at 6:15
    
@KCd, could you expand on what you mean? It seems to me that since $\sqrt{(z-1)(z-2)}$ has a branch point at $\infty$ it can't have a Laurent series there; it's not analytic in any punctured neighborhood of $\infty$. –  Antonio Vargas May 20 '12 at 6:38
    
Strin is asking about "the" Laurent series, but that is not a well-defined concept even for $\sqrt{z}$ at $z=1$. Next I would ask what Strin would say about a Laurent series for $\sqrt{z}$ at $z = 0$, just to avoid involving $\infty$ as part of the issue. –  KCd May 20 '12 at 6:45
    
Ah, I was mistaken. The function does not have a branch point at $\infty$ after all. It turns out that off-by-one errors are kind of important when talking about winding numbers! –  Antonio Vargas May 20 '12 at 7:27
    
@KCd I see, you mean it's not possible to define the Laurent Series at points like $z=0$ for $\sqrt{z}$? –  Strin May 20 '12 at 8:53
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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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But we have to define the branch cut [-$\infty$, 0] in addition to [1,2], for otherwise the function is multi-valued at 0 and discussion of analytic is not allowed. If we defined that branch cut, we cannot say the function is analytic at $\infty$. –  Strin May 20 '12 at 8:51
    
Sorry........$\infty$ is not a branch point and you're right. By the way, it's not possible to define Laurent Series at $z=0$ for $\sqrt{\frac{(t-1)(t-2)}{t}}$ right? –  Strin May 20 '12 at 8:55
    
Right, $\sqrt{(t-1)(t-2)/t}$ has a branch point at $0$ because $\sqrt{1/t}$ does. –  Robert Israel May 20 '12 at 16:07
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