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What is the easiest way to calculate the residue of $\dfrac{\tan(z)}{z^3}$ at zero? I could either use the line integral theorem, or expand it out as a series. Is there a right way to do it?

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up vote 8 down vote accepted

You don't have to calculate the series expansion. The $z^{-1}$-coefficient in the series expansion of $\tan(z)/z^3$ is the $z^2$-coefficient in the series expansion of $\tan(z)$. But this is zero since $\tan$ is an odd function.

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So, more generally, the residue of any even function (at zero) is zero? – GEdgar Jul 31 '12 at 17:15
Yes. For an even function, the Laurent series contains only even powers, so its residue (at zero) is zero. – marlu Jul 31 '12 at 17:24

There is no "right" way to do it, but here is what I would do:

The series at $z=0$ for $\tan(z)=z+\frac{z^3}{3}+\frac{2z^5}{15}+O(z^7)$. Divide that by $z^3$ and look at the coefficient of $\frac1z$.

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$f(z):=\frac{\tan(z)}{z^3}$, $z\neq0$. Then $$ \begin{align} \operatorname{Res}(f,0)&=\frac{1}{2\pi i}\int_{\gamma}f(z)dz\\ &=\frac{1}{2\pi i}\int_{\gamma} \frac{\tan(z)}{z^3}dz\\ &=\frac{1}{2!}\frac{2!}{2\pi i} \int_{\gamma} \frac{\tan(z)}{(z-0)^{2+1}}dz\\ &=\frac{1}{2!}\tan^{(2)}(0)=0, \end{align} $$ where $\tan^{(2)}$ is the second derivative of the $\tan$ function.

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