This is a two part question. First I'm being tasked with the following:
Given $f(z)=\frac{e^{iz}}{z^2+2}$, show that the coefficients for the Taylor expansion with center 0 follow the recursive formula, $a_0=\frac{1}{2}$, $a_1=\frac{i}{2}$ and $a_k = \frac{1}{2}\left(\frac{i^k}{k!}-a_{k-2}\right)$ for $k>2$.
I have no trouble showing $a_0$ and $a_1$, however I have no idea how to show the $k$'th term. Differentiating seems to be an awful mess.
Next I should calculate the following integral:
$$I=\frac{1}{2\pi i}\int_{\partial K\left(\frac{1}{4},\frac{1}{2}\right)}\frac{e^{iz}}{(z^2+2)z^6}dz$$
I think the solution to this is to use Cauchy's integral formula for the $n$'th derivative and show that $a_5=I$. The reason I'm uncertain is because the ball is not centered around 0 but $\frac{1}{4}$. Does that pose a problem or it the solution ($I=a_5$) correct?
If$$f(z)=a_0+a_1z+a_2z^2+a_3z^3+\cdots,$$then\begin{align}1+iz+\frac{i^2}{2!}z^2+\frac{i^3}{3!}z^3+\cdots&=e^{iz}\\&=(z^2+2)f(z)\\&=(z^2+2)\left(a_0+a_1z+a_2z^2+a_3z^3+\cdots\right)\\&=2a_0+2a_1z+(2a_2+a_0)z^2+(2a_3+a_1)z^3+\cdots\end{align}and from this equality you get that$$k\geqslant2\implies 2a_k+a_{k-2}=\frac{i^k}{k!}.$$
Concerning the rest of the problem, you just apply Cauchy's integral formula. The fact that the ball is not centered at $0$ is no relevant. What matters is that you are computing an integral over a loop around $0$.