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## Main Question or Discussion Point

For example, consider:

[tex]f(z)=\sqrt{z(z-1)(z-2)}[/tex]

It's easy to compute the Laurent-Puiseux series in the unit disc, up to the singular point at z=1:

[tex]f(z)=\sqrt{2} \sqrt{z}-\frac{3 z^{3/2}}{2 \sqrt{2}}-\frac{z^{5/2}}{16 \sqrt{2}}-\frac{3 z^{7/2}}{64 \sqrt{2}}-\frac{37 z^{9/2}}{1024 \sqrt{2}}+\cdots,\quad |z|<1[/tex]

That's done by creating a differential equation for the function with polynomial coefficients then solving it using power series where in this case, the indical equation has a root c=1/2 to generate the fractional powers.

But can we generate a Laurent-Puiseux series for the function in the annular region [itex]1<|z|<2[/itex]?

I haven't found any info on the net about this and was hoping someone here could shed some light on the matter.

Thanks,

Jack

[tex]f(z)=\sqrt{z(z-1)(z-2)}[/tex]

It's easy to compute the Laurent-Puiseux series in the unit disc, up to the singular point at z=1:

[tex]f(z)=\sqrt{2} \sqrt{z}-\frac{3 z^{3/2}}{2 \sqrt{2}}-\frac{z^{5/2}}{16 \sqrt{2}}-\frac{3 z^{7/2}}{64 \sqrt{2}}-\frac{37 z^{9/2}}{1024 \sqrt{2}}+\cdots,\quad |z|<1[/tex]

That's done by creating a differential equation for the function with polynomial coefficients then solving it using power series where in this case, the indical equation has a root c=1/2 to generate the fractional powers.

But can we generate a Laurent-Puiseux series for the function in the annular region [itex]1<|z|<2[/itex]?

I haven't found any info on the net about this and was hoping someone here could shed some light on the matter.

Thanks,

Jack