An application of Cauchy integral formula with pole on boundary

Question

How to use Cauchy integral formula to solve the following problem: $$ \oint_{|z-1|=3}\frac{dz}{z(z^2-4)e^z } $$

Answer 1

$$ \oint_{(\text{C})}\frac{dz}{z(z^2-4)e^z}=\oint_{(\text{C})}\frac{dz}{z(z-2)(z+2)e^z} $$

Residue at $z=0$ $$Res_1=\frac{1}{(0-2)(0+2)e^0}=-\frac{1}{4}$$

Residue at $z=-2$ $$Res_2=\frac{1}{(-2)(-2-2)e^{-2}}=\frac{e^2}{8}$$ Since the pole $z=-2$ is turned around of half a spin, the total residue is : $$Res=Res_1+\frac{1}{2}Res_2=-\frac{1}{4}+\frac{1}{2}(\frac{e^2}{8})=\frac{e^2-4}{16}$$ $$ \oint_{(\text{C})}\frac{dz}{z(z^2-4)e^z}=2\pi i Res=\pi i\frac{e^2-4}{8}$$

IN ADDITION , explanation :

On the small semi-circle $(\gamma)\quad\to\quad z=-2+\epsilon\:e^{i\theta}$, the integration leads to : $\oint_{(\gamma)}\frac{dz}{z(z^2-4)e^z}=\int_{\pi/2}^{3\pi/2}\frac{i e^{i\theta}d\theta}{z(z^2-4)e^z}=\frac{1}{2}\int_{\pi/2}^{5\pi/2}\frac{i e^{i\theta}d\theta}{z(z^2-4)e^z}=\frac{1}{2}(2\pi i Res_2)=\pi i Res_2$

The integral on the external contour $(\text{C})+(\gamma)$ is equal to the contribution of the poles inside : $$\oint_{(\text{C})}\frac{dz}{z(z^2-4)e^z}+\pi i Res_2=2\pi iRes_1+2\pi iRes_2$$ Which is the justification of the above result : $$\oint_{(\text{C})}\frac{dz}{z(z^2-4)e^z}=2\pi i\left(Res_1+\frac{1}{2}Res_2\right)$$