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enter image description here I need to find the residue of $\dfrac{1 - \cos z}{z^{3} (z-3)}$ at all its singular points. Is this correct? Also, are there removable singularities? Thanks in advance

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Note that $$\dfrac{1-\cos(z)}{z^3(z-3)} = \dfrac{2 \sin^2(z/2)}{z^3(z-3)} = \dfrac12 \dfrac{\left(\dfrac{\sin^2(z/2)}{(z/2)^2}\right)}{z(z-3)}$$ Can you conclude from this?

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Yes. no removable singular points –  MikeMan May 11 '13 at 19:35

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