Compute complex integral over circle 
Compute the complex integral 
  $$\oint_{C} \frac{e^{z}}{(z+\pi i)^3} dz$$
  where $C$ is the circle with center at $0$ and radius $4$.

Here is my solution, and I hope I can get some correction
let $\gamma: \left[0,2\pi \right] \to \Bbb{C}$
with $\gamma (t)=4e^{it}$. Then 
$\gamma' (t)=4ie^{it}$ and
$$\oint_{C} f(z) dz = \int_{0}^{2\pi} \frac{e^{4e^{it}}}{(4e^{it}+i\pi)^{3}} 4ie^{it} dt.$$
Let $\theta=4e^{it} $, then $ d\theta=4ie^{it}dt$, and 
$$=\int_{0}^{2\pi} \frac{e^{\theta}}{(\theta+i\pi)^{3}} 4ie^{it} \frac {d\theta}{4ie^{it}}=
\int_{0}^{2\pi} \frac{e^{\theta}}{(\theta+i\pi)^{3}} d\theta$$
Now I'm stuck here. Did I make some mistakes?
 A: I do not suggest to evaluate this integral directly.
Are you familiar with the Residue Theorem?
In that case, since $|-\pi i|<4$, it follows that
$$I:=\oint_{|z|=4} \frac{e^{z}}{(z+\pi i)^3} dz=2\pi i\mbox{Res}\left(\frac{e^{z}}{(z+\pi i)^3},z=-\pi i\right).$$
Since the pole has order three then (see link)
$$\mbox{Res}\left(\frac{e^{z}}{(z+\pi i)^3},z=-\pi i\right)=
\frac{1}{2!}\lim_{z\to -\pi i} (e^{z})''=\frac{e^{-\pi i}}{2}=-\frac{1}{2}.$$
Therefore $I=-\pi i$.
Another approach. By  Cauchy Theorem, the integral is equal to
$$I:=\oint_{|z+\pi |=r} \frac{e^{z}}{(z+\pi i)^3} dz=\oint_{|w|=r} \frac{e^{w-\pi i}}{w^3} dw=-\oint_{|w|=r} \frac{1+w+\frac{w^2}{2}+w^3h(w)}{w^3} dw$$
where $r>0$ and $h$ is holomorphic in $\mathbb{C}$. Again by Cauchy Theorem, $\oint_{|w|=r}h(w)dw=0$, hence
$$I:=-\oint_{|w|=r} \frac{dw}{w^3}-\oint_{|w|=r} \frac{dw}{w^2}-\frac{1}{2}\oint_{|w|=r} \frac{dw}{w}=0+0-\pi i=-\pi i$$
where in the last step we used the substitution $w=re^{it}$ with $t\in[0,2\pi]$.
P.S. BTW your substitution $[0,2\pi]\ni\theta=4e^{it}$ is not correct.
A: Hint: Use Cauchy's integral formula:
$$
f^{(n)}(a) = \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{(z-a)^{n+1}}\, dz
$$
with $f=\exp$, $a=-\pi i$, $n=2$.
