$\bf Exercise:$ Determine the residuals in all isolated singularities of the function $ f (z) = \tan ^ 3 (z) $
$\bf \text{Idea}:$ Note that $$f(z)= \tan ^ 3 (z)= \frac{\sin ^ 3 (z)}{\cos ^ 3 (z)}$$ and $$\cos ^ 3 (z)=0 \Rightarrow z=k\pi+\frac{\pi}{2}, k\in\mathbb{Z}$$
Then $z=k\pi+\frac{\pi}{2}$ they are poles of order $3$. Hence we must calculate the residuals in $z=k\pi+\frac{\pi}{2}$,one way to do this is to calculate the Laurent series of $f(z)$ around of $z=k\pi+\frac{\pi}{2}$. For convenience in accounts, be $z=u+k\pi+\frac{\pi}{2}$ then:
$\frac{\sin ^ 3 (z)}{\cos ^ 3 (z)}=\frac{\sin ^ 3 (u+k\pi+\frac{\pi}{2})}{\cos ^ 3 (u+k\pi+\frac{\pi}{2})}=\frac{(\sin(u+k\pi)\cos(\frac{\pi}{2})+\sin(\frac{\pi}{2})\cos(u+k\pi))^3}{(\cos(u+k\pi)\cos(\frac{\pi}{2})-\sin(u+k\pi)\sin(\frac{\pi}{2}))^3}=\frac{(-1)^{3k}\cos^3(u)}{(- 1) ^ {3k+1}\sin ^ 3 (u)}=-\frac {\cos ^ 3 (u)} {\sin ^ 3 (u)}$
Therefore, we must calculate the Laurent series of $ -\frac {\cos ^ 3 (u)} {\sin ^ 3 (u)} $ around $ u = 0 $, and look for the coefficient of $ \frac {1} {u}$.
But how can I do this? Is my idea correct? Is there another shorter way to solve the problem? Thanks in advance for the suggestions.
Near $\frac\pi2$, you have$$\tan(z)=\frac1{z-\frac\pi2}\left(-1+\frac13\left(z-\frac\pi2\right)^2+\frac1{45}\left(z-\frac\pi2\right)^4+\cdots\right)$$(actually, the $\frac1{45}\left(z-\frac\pi2\right)^4$ term will not be needed) and therefore\begin{align}\tan^3(z)&=\frac1{\left(z-\frac\pi2\right)^3}\left(-1+\frac13\left(z-\frac\pi2\right)^2+\frac1{45}\left(z-\frac\pi2\right)^4+\cdots\right)^3\\&=\frac1{\left(z-\frac\pi2\right)^3}\left(-1+\left(z-\frac\pi2\right)^2-\frac4{15}\left(z-\frac\pi2\right)^4+\cdots\right)\\&=-\frac1{\left(z-\frac\pi2\right)^3}+\frac1{\left(z-\frac\pi2\right)}-\frac4{15}\left(z-\frac\pi2\right)+\cdots\end{align}and therefore $\operatorname{res}_{z=\pi/2}\tan^3(z)=1$. The same argument works at the other poles.