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Is a pole of fractional order just another name for a certain type of branch point?

The function $\frac{1}{\sqrt{z}}$ has a logarithmic algebraic branch point at $z=0$.

But according to Wolfram Alpha, $ \frac{1}{\sqrt{z-1}}$ has a pole of order $\frac{1}{2}$ at $z=1$.

EDIT:

When asked, Wolfram Alpha still says that $\frac{1}{\sqrt{z-1}}$ has a pole of order $\frac{1}{2}$ at $z=1$.

But now at least it also says that $\frac{1}{\sqrt{z-1}}$ has a branch point there.

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The branch point of $\frac 1{\sqrt z}$ is not logarithmic according to Wikipedia's definition, since if you go two times around the branch point, you end up in the branch you started with. In any case, whichever terminology you choose, the two functions are just translates of each other and therefore their singularities at 0 resp 1 are of the same kind. –  Henning Makholm Mar 23 '12 at 20:43

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For an illustration of the two leaves of the Riemann surface of the square root see Wikipedia. For $f(z)=\sqrt[3]{z}$ you would have three leaves and so on...

You may compare this to the logarithmic branches of $\log$. (a comparison is at the second page of this paper).

But you wanted the multiplicative inverse of the square root $f(z)=\frac 1{\sqrt{z}}$. An illustration of the imaginary part $z=\Im\bigl(\frac 1{\sqrt{x+iy}}\bigr)$ is :

As explained by Henning Makholm $f(z)=\frac 1{\sqrt{z-1}}$ is merely the function $f(z)=\frac 1{\sqrt{z}}$ with center at $0$ shifted at $1$.

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There is no such thing as a pole of fractional order. $1/\sqrt{z}$ and $1/\sqrt{z-1}$ have algebraic, not logarithmic, branch points.

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"Pole of fractional order" has only 457 matches in Google. As opposed to more standard concepts with thousands or millions of matches. –  GEdgar Mar 23 '12 at 23:57

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