Q:

Radius of convergence of the taylor series of $f(z)=\frac{e^{iz}}{z^2+2}$ about $z=0$

Let $f(z)=\frac{e^{iz}}{z^2+2}$
I need to determine the radius of convergence of the taylor series of $f$ about $z=0$.
My 1st approach
If I can write $f$ as a power series, then it will be equal to the taylor series about 0, and thus I can determine the radius by using the cauchy-hadamard theorem. However, I haven't been able to write $f$ as a power series.
My 2nd approach
Using the cauchy integral formula, I can determine the taylor series directly:
$$
\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n =
\sum_{n=0}^{\infty} \frac{1}{2\pi i} \int_{\partial K(0,r)}
\frac{f(z)}{z^{n+1}} dz \cdot z^n
$$
where $\partial K(0,r)$ is some appropriate circle with center 0. However, this seems to be difficult to evaluate...
My 3rd approach
We can write $f$ as a product of two series:
$$
f(z)=\frac{e^{iz}}{z^2+2}=\frac{1}{2} \frac{1}{1-(-\frac{z^2}{2})} e^{iz}
    =\frac{1}{2} \sum_{n=0}^{\infty} \left(-\frac{z^2}{2}\right)^n
     \cdot \sum_{n=0}^{\infty} \frac{(iz)^{n}}{n!}
$$
and since the 2nd infinite series has infinite radius of convergence, we only need to determine the radius of convergence of the first geometric series. However, I don't know how this is related to the taylor series of $f$...
Help would be greatly appreciated!

A:

The power series of a holomorphic function converges on the largest disc contained in the domain. Now the function is holomorphic on $\mathbb{C}\setminus \{i\sqrt{2},−i\sqrt{2}\}$, so it will converge on a disc with radius $\sqrt{2}$. 
To check that it does not converge on a larger disc, we must check that $i\sqrt{2}$ and $−i\sqrt{2}$ are not removable singularities, which they are not. 

