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i already ask that one, but i am now confused, i been trying using series for the first and second one but i really don´t know how to continue so i need help and a rigorous, formal and step by step answer. Like always i really appreciate your help.

$$\int_\gamma e^{\dfrac{1}{z^2}}dz$$ $$\int_\gamma e^{\dfrac{1}{z}}dz$$

where $$\gamma:|z|=1 $$ $$\int_{-\pi}^\pi \frac{d\theta}{1+sin^2(\theta)}$$ Thanks

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up vote 1 down vote accepted

The function $e^x$ has the Maclaurin series $$1 + x + \frac{x^2}{2!} + \mathcal{O}(x^3)$$ if we make the substitution $x\mapsto \frac{1}{z^2}$ then we get the Laurent series $$e^\frac{1}{z^2} = 1 + \frac{1}{z^2} + \frac{1}{2!\,z^4} + \mathcal{O}\left(\frac{1}{z^6}\right)$$ The residue for the above series is the coefficient of the term $\frac{1}{z}$. That term does not appear in the Laurent series, so it is simply $0$.

If we instead make the substitution $x\mapsto \frac{1}{z}$ then we have the Laurent series $$e^\frac{1}{z} = 1 + \frac{1}{z} + \frac{1}{2!\,z^2} + \mathcal{O}\left(\frac{1}{z^3}\right)$$ the residue in this case is then $1$.

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