$\oint_C \frac{\sin(z)}{e^z z^3}dz$

Question

Evaluate the given integral $$ \oint_{C}\frac{\sin\left(\,{z}\,\right)}{{\rm e}^{z}z^{3}} \,{\rm d}z,$$ where $C$ is the circle $\left\vert\,{z - 1}\,\right\vert = 3$.

I'm stuck, I don't know which integration method to use, any hint, please $?$ :(

Answer 1

Using Cauchy's formula: $$f''(z_0)= \frac{2!}{2\pi i}\oint_{\partial D} \frac{f(z)}{(z-z_0)^3}d z,$$ where $f(z)$ requires to be holomorphic inside domain $D$.

In your case take $f(z) = \frac{\sin z}{e^z}$ and $z_0=0$. $$f'(z) = \frac{\cos z - \sin z }{e^{z}}$$ and $$f''(z) = \frac{-\sin z - \cos z - \cos z + \sin z}{e^z} = \frac{-2\cos z}{e^z}.$$ So the integral is $\frac{2\pi i}{2} f''(0) = -2\pi i $