Use cauchy's integral formula to evaluate

$\displaystyle \oint \frac{e^{sin(z)}}{z^2(z-\frac{\pi}{4})}$

over the curve,

$\displaystyle c:=|z|=\frac{\pi}{8})$

Okay, so i used the method shown in lectures,

Since the only singularity inside C is for z=0,

we let ,

$\displaystyle f(z)=\frac{e^{sin(z)}}{(z-\frac{\pi}{4})}$

hence, by cauchy's formula, z=0

$\displaystyle (2\pi i).\frac{e^{sin(0)}}{(0-\frac{\pi}{4})}= -8i$

But the answer says -32/pi*(1+pi/4)i

Just starting this topic, so still trying to wrap my head around it.

Thanks in advanced