Radius convergence of $f(z) = \frac{1}{1+\sin^2(z)}$ Determine the radius convergence of Taylor's series for the following function: $$f(z) = \frac{1}{1+\sin^2(z)}$$ around  $z_0 = 0$.
Now here is what I have tried to do:
$\sin^2(z) = 1 - \cos^2(z)$ , $\cos^2(z) = \frac{1}{2}(\cos(2z) + 1)$
so the last form would be $\frac{2}{1-\cos(2z)}$
, we know that $$\cos(z) = \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{(2n)!}$$
then
$\cos(2z) = \sum_{n=0}^{\infty} (-1)^n \frac{(2z)^{2n}}{(2n)!}$ ,
So the last form would be $\frac{2}{1-\ \sum_{n=0}^{\infty} (-1)^n \frac{(2z)^{2n}}{(2n)!})}$
which is $\frac{2}{-\ \sum_{n=1}^{\infty} (-1)^n \frac{(2z)^{2n}}{(2n)!}}$
And I am stuck here , I have no idea what to do next
If for f(z) a finite number of singular points in the complex plane, then the radius of convergence R is
The distance from z0 to the nearest singular point.
 A: The radius of convergence is the distance from $0$ to the nearest non-removable singularity. And we have\begin{align}1+\sin^2(z)=0&\iff\sin(z)=\pm i\\&\iff\frac{e^{iz}-e^{-iz}}{2i}=\pm i\\&\iff e^{iz}-e^{-iz}=\mp2\\&\iff\left(e^{iz}\right)^2\pm2e^{iz}-1=0\\&\iff e^{iz}=1+\sqrt2\vee e^{iz}=1-\sqrt2\vee e^{iz}=-1+\sqrt2\vee e^{iz}=-1-\sqrt2\\&\iff iz\in\log\left(1+\sqrt2\right)+2\pi i\Bbb Z\vee iz\in\log\left(\sqrt2-1\right)+\pi i+2\pi i\Bbb Z\vee{}\\&\qquad\vee iz\in\log\left(-1+\sqrt2\right)+2\pi i\Bbb Z\vee iz\in\log\left(1+\sqrt2\right)+\pi i+2\pi i\Bbb Z\\&\iff z\in i\log\left(1+\sqrt2\right)+2\pi\Bbb Z\vee z\in\pi+i\log\left(\sqrt2-1\right)+2\pi\Bbb Z\vee{}\\&\qquad\vee z\in i\log\left(-1+\sqrt2\right)+2\pi\Bbb Z\vee z\in\pi+i\log\left(1+\sqrt2\right)+2\pi\Bbb Z.\end{align}Of all these numbers, the ones which are closer to $0$ are $i\log\left(\pm1+\sqrt2\right)$. Therefore, the radius of convergence is $\bigl|\log\left(\pm1+\sqrt2\right)\bigr|\approx0.88$.
A: If a power series $P(z) = \sum a_nz^n$ converges on an open disc $\{z : |z| < r\}$ and this function $P(z)$ has an analytic continuation to all points on the circle $\{z : |z| = r\}$, then (by using compactness of $\{z : |z| = r\}$) the original power series $\sum a_nz^n$ converges on a bigger open disc $\{z : |z| < r+\varepsilon\}$, so the function $P(z)$ defined by that original power series has an analytic continuation to that bigger open disc.
Theorem. If $f(z)$ is a meromorphic function on $\mathbf C$ that is analytic at $0$ $($so it has no poles on some open disc around $0$$)$, then the radius of convergence of the power series for $f(z)$ at $0$ is the distance from $0$ to the nearest pole for $f(z)$.
Proof.  Let $R$ be the distance from $0$ to the nearest pole of $f(z)$, so $f$ is analytic on $\{z : |z| < R\}$.  Let the power series for $f(z)$ at $0$ have radius of convergence $r$.
We show $r \leq R$ by contradiction. If $r > R$ then that series would be analytic on $\{z : |z| < r\}$, so it must agree with $f$ on  $\{z : |z| < r\}$ (uniqueness of meromorphic continuation), but $f$ is not analytic on that disc because of its pole with absolute value $R$, which is less than $r$.
Suppose $r < R$. Since $f$ is analytic on $\{z : |z| < R\}$ while its power series at $0$ converges on the smaller disc $\{z : |z| < r\}$ and is analytic on all $z$ with $|z| = r$, that power series at $0$ converges on a bigger disc $\{z : |z| < r + \varepsilon\}$, so the radius of convergence of that series is bigger than $r$.
Thus $r = R$. $\Box$
Apply this to $1/(1 + \sin^2 z)$, which is meromorphic on $\mathbf C$ with poles precisely where $\sin^2 z = 1$, which is equivalence to
$\sin z = \pm i$, so you just need to solve $\sin z = i$ and $\sin z = -i$ and find the smallest solution. Since $\sin z = -i$ is equivalent to $\sin \overline{z} = i$ (because $\overline{\sin z} = \sin \overline{z}$), the solutions to $\sin z = i$ and $\sin z = -i$ come in conjugate pairs, so it suffices to look at solutions of $\sin z = i$:
\begin{eqnarray*}
\frac{e^{iz} - e^{-iz}}{2i} = i & \Longleftrightarrow & e^{iz} - e^{-iz} = -2 \\
&  \Longleftrightarrow & e^{iz}  + 2 - e^{-iz} = 0 \\
&  \Longleftrightarrow & e^{2iz} + 2e^{iz}  - 1 = 0.
\end{eqnarray*}
The roots of $w^2 + 2w - 1 = 0$ are $w = (-2 \pm \sqrt{8})/2 = -1 \pm \sqrt{2}$, so we need to solve
$$
e^{iz} = -1 + \sqrt{2} \ \ {\rm and } \ \ e^{iz} = -1-\sqrt{2}.
$$
and find the solution $z$ where $|z|$ is minimal.  The other answer discusses that.
