Complex var. integral: $\oint_{|z|=1} \frac{z^2\ dz}{\sin^3{z}\cos{z}}$

Question

Integrate $\displaystyle\oint_C \dfrac{z^2\ dz}{\sin^3{z}\cos{z}}$; $C \rightarrow |z|=1$

I already know that $|z|=1$ is a circumference with $r=1$ and center at $(0,0)$.

I also know there are three methods using remainders to solve this kind of integrals. But I don't know which method I should use:

1. Using Laurent's series (consider $a_{-1}$), so $\displaystyle\oint_C f(z) dz = 2\pi i \cdot a_{-1}$
2. $\displaystyle\text{Rem}\ f(z_0) = \left[ \frac{1}{(n-1)!} \right] \lim_{z\to0}\frac{d^{n-1}}{dz^{n-1}}\left[f(z) (z-z_0)^n\right]$, when $z$ is a $n$-th grade pole.
3. If $\displaystyle f(z) = \frac{\varphi(z)}{\psi(z)}$, $\varphi, \psi$ analythical, and $\varphi(z_0) \neq 0$, $\psi(z_0) = 0$, $\psi'(z_0) \neq 0$, then $\text{Rem}\ f(z_0) = \frac{\varphi(z_0)}{\psi'(z_0)}$

Thanks in advance =)

Answer 1

There is a simple pole at the origin ($z^2/\sin^2{z}$ is defined to be $1$ at the origin). No other pole of the integrand is within $C$ so by the residue theorem, the integral is $i 2 \pi$.