The radius of convergence of series $\frac{1}{2+\sin z}$ at $z = 0$

Question

What is the radius of convergence for this series at $z = 0$?
I've tried this:
As it was said in the comments, $\sin z = -2$, from which I've got
$1.\ z = 2\pi n - \sin^{-1}(2), \qquad n\in \mathbb Z$
$2.\ z = 2\pi n + \pi + \sin^{-1}(2), \qquad n \in \mathbb Z$

Answer 1

As hinted by @MPW in the first comment, the radius of convergence is the distance between the origin (the "center of the expansion") and the nearest $z$ such that $\sin z=-2.$ So, it is the minimum possible absolute value of $z$ such that $e^{iz}-e^{-iz}=-4i.$

Let $Z:=e^{iz}=e^{ix-y}$ $$\begin{align}Z-\frac1Z=-4i&\iff Z^2+4iZ-1=0\\&\iff(Z+2i)^2=-3\\&\iff Z=-i(2\pm\sqrt3)\\&\iff-y=\ln(2\pm\sqrt3),\;x=-\frac\pi2+2k\pi\;(k\in\Bbb Z). \end{align}$$

The radius of convergence is therefore $$R=\sqrt{\frac{\pi^2}4+\ln^2(2-\sqrt3)}.$$

Answer 2

You can re-write your function this way: $$ \mathrm{sin}(z) = \frac{e^{iz} - e^{-iz}}{2i}, \quad 2 + \sin(z) = 2 + \frac{e^{iz} - e^{-iz}}{2i} = \frac{e^{iz} + 4i - e^{-iz}}{2i} $$ The function $e^{iz}$ has no zeros or poles, so multiplying it doesn't change the location of the zeros and poles (except at infinity, but we are currently not interested in that). We see that $$ 2i e^{iz} (2 + \sin(z)) = e^{2iz} + 4i e^{iz} - 1 $$ and the latter is equal to zero if and only if $e^{iz}$ is a root of the quadratic equation $$ x^2 + 4i x - 1 = 0 $$ and by the quadratic formula, this gives $$ x = \frac{-4i \pm \sqrt{(4i)^2 - 4(1)(-1)}}2 = -2i \pm \frac{\sqrt{-16 + 4}}2 = -2i \pm \sqrt 3. $$ To solve $e^{iz} = a + bi$, we first write $z = x + iy$ with $x,y$ real and then note that $$ e^{iz} = e^{i(x+iy)} = e^{-y + ix} = e^{-y} e^{ix}. $$ Writing $a + bi$ in polar form (i.e. $a + bi = \sqrt{a^2 + b^2} e^{i \theta(a,b)}$), we find $$ e^{-y} = \sqrt{a^2 + b^2} \quad \Rightarrow \quad y = -\mathrm{ln}(\sqrt{a^2 + b^2}). $$ We also find $\cos(\theta) = \frac a{\sqrt{a^2 + b^2}}$ and $\sin(\theta) = \frac b{\sqrt{a^2 + b^2}}$, which allows figuring out the angle.

In our case, we have $y = -\mathrm{ln}(\sqrt 7)$ by the above formula. For $x$, we have $\sqrt{a^2 + b^2} = \sqrt 7$ in both cases since the square kills the sign change. We also have $$ \cos(\theta) = \pm \sqrt{3/7}, \quad \sin(\theta) = \frac{-2}{\sqrt 7}. $$ We are looking for the closest value of $z$ to the origin for which $e^{iz}$ solves the previous quadratic equation, and since $\theta(a,b)$ can be replaced by an integer multiple of $2\pi$, we need to find the closest value it can take to $0$ in the interval $]-\pi, \pi]$ and assign that to $x$; our answer for the radius of convergence will become $\sqrt{x^2 + y^2}$. We get

$$ \tan(\theta) = \mp \frac 2{\sqrt 3} $$ which after computing $\tan^{-1}$ numerically gives a value of approximately $0.857$ in absolute value for $\theta$. We need to check signs to confirm the actual angles though.

In the case where $\cos(\theta) = +\sqrt{3/7}$, that puts our angle in the fourth quadrant. So since the value of $\tan(\theta)$ is $-\frac 2{\sqrt 3}$, we see that $\theta(\sqrt 3, -2) \approx -0.857$.

In the case where $\cos(\theta) = -\sqrt{3/7}$, that puts our angle in the third quadrant with $\tan(\theta) = + \frac 2{\sqrt 3}$. Therefore, to fit our angle in the interval $]-\pi, \pi]$, we need to subtract $\pi$ from it, so $0.857$ becomes something that is roughly $-2.2845$. Clearly our first value is correct, so $\theta(-\sqrt 3, -2) = \tan^{-1}(-\frac 2{\sqrt 3}) \approx -0.857$.

Our radius of convergence is therefore $$ \sqrt{(\tan^{-1}(-2/\sqrt 3))^2 + (-\mathrm{ln}(\sqrt 7))^2} = \sqrt{(\tan^{-1}(-2/\sqrt 3))^2 + (\mathrm{ln}(\sqrt 7))^2}. $$

I gave the rough idea, and there might be more than one approach. I apologize if I made any computational errors; feel free to point them out if you find some, and I will adjust my answer.

Hope that helps,