Calculate $\frac{1}{2\pi i} \int_{\gamma} \frac{e^{iz}}{z^6(z^2+2)} dz$, where $\gamma$ is the circle with center 1/4 and radius 1/2 I need to calculate the following complex integral:
$$
\frac{1}{2\pi i} \int_{\gamma} \frac{e^{iz}}{z^6(z^2+2)} dz
$$
where $\gamma$ is the circle with center $1/4$ and radius $1/2$. It seems natural to use Cauchy's Integral Formula, since we can split the integrand into the two terms,
$$
\frac{e^{iz}}{z^6(z^2+2)}
=\frac{e^{iz}}{2\sqrt{2}iz^6(z-\sqrt{2}i)}
-\frac{e^{iz}}{2\sqrt{2}iz^6(z+\sqrt{2}i)}
$$
but before using the formula, somehow the $z^6$ in both denominators has to be dealt with. I've tried splitting up the terms once again, but without any succes so far. I've also tried to insert $e^{iz}=\sum \frac{(iz)^n}{n!}$ and interchange integral and sum, this removes $z^6$ from both denominators, but the resulting series, I think, cannot be calculated exactly.
Help would be appreciated.

Without residues you may proceed as follows:
The function $f(z) = \frac{e^{iz}}{z^2+2}$ is holomorph on your disc and contains $z= 0$.
So, Cauchy integral formula gives
$$f^{(5)}(0) = \frac{5!}{2\pi i }\oint_{\gamma}\frac{\frac{e^{iz}}{z^2+2}}{z^6}dz$$
Now, you know that $\frac{f^{(5)}(0)}{5!}$ is the coefficient of $z^5$ in the McLaurin-expansion of $f$
\begin{eqnarray*} [z^5]\frac{e^{iz}}{z^2+2}
& = & [z^5]\frac{1}{2}\frac{e^{iz}}{1 - \left(-\frac{z^2}{2}\right)} \\
& = & [z^5]\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{(iz)^n}{n!}\right)\left(\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n}}{2^n}\right) \\
& = & \frac{1}{2}\left( \frac{i^5}{5!} + \frac{i^3}{3!}\left(-\frac{1}{2}\right)+\frac{i}{4}\right) \\
& = & \boxed{\frac{41}{240}i}
\end{eqnarray*}
