Residue of $\sin(\tan(z))$ at $z=\frac{\pi}{2}$

Question

In finding this residue, I took a Taylor expansion of $\sin(x)$ and substituted in $\tan(z)$ and then attempted to compute the residues of the odd powers of $\tan(z)$ at $z=\frac{\pi}{2}$. Using Wolfram Alpha, it seemed to be that: \begin{equation} \operatorname{Res}(\tan^{2n-1}(z), \frac{\pi}{2}) = (-1)^n \end{equation} Giving the residue of $\sin(\tan(z))$ as $-\sinh(1)$.

Is this correct?

Also, I was unable to show $\operatorname{Res}(\tan^{2n-1}(z), \frac{\pi}{2}) = (-1)^n$.

Is this true and if so any hints on how to show it?

Answer 1

Using $\sin\tan z=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)!}\left(\tan z\right)^{2n+1}$ is a good idea. Then $$ \operatorname*{Res}_{z=\pi/2}\left(\tan z\right)^{2n+1}= -\operatorname*{Res}_{z=0}\left(\cot z\right)^{2n+1}=-\frac{1}{2\pi i}\oint_{\|z\|=\varepsilon}\left(\cot z\right)^{2n+1}\,dz$$ by Cauchy's integral formula. The substitution $z=\arctan u$ allows to write the RHS as $$ -\frac{1}{2\pi i}\oint_{\|u\|=\varepsilon}\frac{du}{(u^2+1)u^{2n+1}}=-\operatorname*{Res}_{u=0}\frac{1}{u^{2n+1}}\left(1-u^2+u^4-u^6+\ldots\right) $$ hence $$ \operatorname*{Res}_{z=\pi/2}\left(\tan z\right)^{2n+1}= (-1)^{n+1}, $$ $$\operatorname*{Res}_{z=\pi/2}\sin\tan z=-\sum_{n\geq 0}\frac{1}{(2n+1)!}=-\sinh(1).$$

Answer 2

As a complement to Jack's answer, we can write for any $\epsilon \in (0,\pi/2)$

$$\begin{align} \oint_{|z-\pi/2|=\epsilon}\sin(\tan(z))\,dz&\overbrace{=}^{z\mapsto z+\pi/2}-\oint_{|z|=\epsilon}\sin(\cot(z))\,dz\\\\ &\overbrace{=}^{z\mapsto \arctan(z)}-\oint_{|z|=\tan(\epsilon)}\frac{\sin(1/z)}{1+z^2}\,dz\\\\ &=-\oint_{|z|=\tan(\epsilon)}\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)!z^{2n+1}}\,\sum_{m=0}^\infty (-1)^mz^{2m}\,dz\\\\ &=-\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)!}\,\sum_{m=0}^\infty (-1)^m\oint_{|z|=\tan(\epsilon)}z^{2m-2n-1}\,dz\\\\ &=-\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)!}\,\sum_{m=0}^\infty (-1)^m 2\pi i \delta_{mn}\\\\ &=-2\pi i \sum_{n=0}\frac{1}{(2n+1)!}\\\\ &=2\pi i (-\sinh(1)) \end{align}$$

as expected!