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Find all singularities of $ \ \frac{z-1}{z^2 \sin z} \ $

Determine if they are isolated or nonisolated.

This is not hard, it is z = 0 and z = k*pi.

But how do I:

For isolated singularities, determine if they are removable or nonremovable and, if nonremovable, determine their order.

Do I need to expand this to a power series? If so, I have no idea where to attack...

Please help! Thanks!

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up vote 2 down vote accepted

Since for any $\,k\in\Bbb Z-\{0\}\,$ we have that

$$\lim_{z\to k\pi}(z-k\pi)\frac{z-1}{z^2\sin z}\stackrel{\text{L'Hospital}}=\lim_{z\to k\pi}\frac{z-1}{2z\sin z+z^2\cos z}=$$

$$=\frac{k\pi -1}{2k\pi \sin k\pi+k^2\pi^2\cos k\pi}=\frac{(-1)^k(k\pi -1)}{k^2\pi^2}$$

all these singularities are isolated (in fact, simple poles).

About the case $\,k=0\,$:

$$\lim_{z\to 0}z^3\frac{z-1}{z^2\sin z}=\lim_{z\to 0}\frac{z}{\sin z}(z-1)=-1$$

Thus we have here a pole of order $\,3\,$

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