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I am trying to find the type of singularity the function $$f(z) = \exp\left(\frac{(\cos z-1)^2}{z^4}\right).$$ has at $z = 0$.

The expression reduces to $$\exp\left(\frac{\sin^4(z/2)}{z^4/2}\right).$$ The function has removable singularity at $z=0$ as $$\lim_{z\rightarrow 0}zf(z)=0.$$

Am I right?

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    The expression you say the original one reduces to is incorrect.2012-07-29

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