# 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!

• If i remember correctly, then 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 $\sqrt{2}$. – Leander Tilsted Kristensen Sep 25 '19 at 20:00
• If an analytic function has the first singularity (in absolute value) at $|z_0|=R$ then the radius of convergence of the Taylor series at $0$ is precisely $R$ so here this gives $\sqrt 2$ as the answer. To prove the above use cauchy estimates for the Taylor series coefficients on any smaller disc $r <R$ to show the Taylor series has radius at least $r$ and then conclude since the radius clearly cannot be bigger than $R$ – Conrad Sep 25 '19 at 20:05
• Of course, I should have seen that. The einstellung effect, I guess... – Tom Sep 25 '19 at 20:09
• @LeanderTilstedKristensen Why not an official answer? – Paul Frost Sep 26 '19 at 9:52

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.